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I have been doing some reading about \$I_{CEO}\$ and \$I_{CBO}\$, but couldn't really understand how \$I_{CBO}\$ is getting amplified (1+ \$\beta\$) times to reach \$I_{CEO}\$, in the open base condition.

This is my understanding so far.... The \$I_{CBO}\$ flows from collector to base. You talk of \$I_{CEO}\$ when the base terminal is open. The \$I_{CBO}\$ will actually have to flow through the emitter to reach back to the collector. The textbooks say, now the \$I_{CBO}\$ also gets amplified "just like \$I_B\$", to get a bigger value: \$I_{CEO}\$.

But our transistor now has only 2 terminals. How does the transistor amplify the \$I_{CBO}\$ to \$I_{CEO}\$ with just two terminals?

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  • \$\begingroup\$ Reverse junction collector-base current is just like regular base current - why shouldn't it get amplified like regular base current - it's exactly the same for a phototransistor - it doesn't have a base (some do maybe) - it's the variation in reverse current due to light variations that gets amplified. All that is important is that there is an equivalent base current I would have thought. \$\endgroup\$
    – Andy aka
    Commented Aug 10, 2014 at 10:43
  • \$\begingroup\$ Ah, I guess phototransistor too need an external stimulus(in the form of light) to be applied in the base, to control a large current flow between its terminals. But here there is no external stimulus. the temp and all is considered constant. only the base terminal is open. \$\endgroup\$
    – ibsen
    Commented Aug 10, 2014 at 12:12
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    \$\begingroup\$ While ever temperature isn't -273.14 degrees C there is stimulus \$\endgroup\$
    – Andy aka
    Commented Aug 10, 2014 at 14:22

2 Answers 2

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I think you can't answer this with currents as above, instead you have to use charge flows, i.e. the drift of specific carrier populations.

Let's take an NPN transistor for example, with a supply connected between C and E to reverse-bias the CB junction.

At the micro level, what is ICBO composed of? Currents in reverse biased diodes are from random thermal creation of hole/electron pairs, so in this case we'd find new electrons being pulled out of the CB depletion zone and flowing to the collector, and new holes being pulled the other way and flowing into the base. Holes into the Base! There aren't electrons in that P-doped region to swallow the incoming holes, so they can diffuse through to the BE junction, meeting up with electrons, and reducing the BE depletion layer thickness, so forward-biasing that junction. (So, essentially the CB diode acts like a large resistor in series with the BE junction.)

But when the BE depletion zone narrows, the lowered barrier allows the vast population of mobile electrons in the Emitter to start crossing the junction into the Base region. So: small hole-flow from CB is turning on Vbe, causing large electron-flow coming from E. (This involves hfe, just as when Base is connected, where Ib produces a small hole-flow into BE junction, lowering Vbe and producing very large electron current Ie ~ hfe * Ib.)

Carriers flooding across BE junction. Electrons in the Base! Some will be swallowed by holes, but since the Base is thin and lighter-doped, most will diffuse across. When electrons arrive at the reverse-biased CB junction they're yanked across by the large e-field there.

Wouldn't this produce runaway, hfe*hfe*hfe...? Nope, and neither does runaway happen in diodes: forward biasing DOESN'T trigger even more forward biasing and so a runaway. It's because the BE junction is being forward biased by the holes (not electrons) coming from the Base region. The large amount of electrons pouring in from the Emitter region don't turn on the BE diode any more than it already is. Or in other words, Icbo is a hole-current just like Ib would be, and the large Ie doesn't reverse course while flipping its carrier polarity to make Icbo larger. Icbo and Ie are separate clouds of carriers flowing in the same piece of silicon. Icbo can "communicate with" the Ie via the BE junction-barrier and Vbe. But the two don't interact directly, in the simple case.

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  • \$\begingroup\$ Let's take an NPN transistor for example, with a supply connected between C and E to reverse-bias the CB junction. How is it possible to make CB junction reverse at first when B is floatig? \$\endgroup\$
    – emnha
    Commented Nov 17, 2021 at 18:40
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    \$\begingroup\$ @emnha the B isn't perfectly floating (not infinite ohms.) A reverse-biased diode usually looks like ten Megs, or maybe 100, so with B disconnected the EC current is perhaps 1uA or less. The silicon is full of mobile carriers, and if we connect an EC supply, the voltage-fields will push these carriers around. In the forward-biased EB diode, holes and electrons are pushed together, turning the diode on. In the reversed CB diode, holes and electrons are pulled away from each other, so the depletion zone widens (and it behaves as a large-value resistance, not a perfect insulator.) \$\endgroup\$
    – wbeaty
    Commented Nov 19, 2021 at 4:38
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ICEO and ICBO exist/or measured when you connect the transistor in two different configuration(common base and common emmiter respectively) so there is no question of asking whether both exist simultaneously(answer is no).Consider you have transistor say T1 to be measured.You measure both Icbo and Iceo when you connect T1 in two different config as mentioned above.So now when you want to know what the value of Icbo wrt measured Iceo Iceo=(1+beta)Icbo will be the expression for solving.In a nutshell to calculate either Icbo or Iceo you need to know to measure just either of Iceo or Icbo for which the expression helps.Hope you get it! Cheers

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