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Another form of the question is: will joining two diodes with wires (pn-np) make a transistor-equivalent?

I read that they are not equivalent, but why?

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    \$\begingroup\$ For one thing, holes won't go through metal wires. \$\endgroup\$ – Dmitry Grigoryev Mar 27 '17 at 7:40
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Many people think that the answer to this question is related to the width of Base region in BJT transistors - it is incorrect. The answer got quite long. You can read starting from "Tricky Question" section if you want the bottom line.

I believe that you were led to ask this question due to something like this picture:

enter image description here

This is a standard practice of teaching the basics of BJT, but it can confuse someone not familiar with semiconductors theory in details.

In order to answer your question at an acceptable level, I need to assume that you're familiar with the principles of operation of PN diode. This reference contains a detailed discussion of PN junctions.

The answer concerns NPN transistor, but it also applies to PNP transistors after appropriate change of polarities.

enter image description here

NPN in forward-active mode of operation:

The most "useful" mode of operation of BJT transistor is called "forward-active":

enter image description here

NPN is in forward-active mode when:

  • Base-Emitter junction is forward biased (usually at \$V_{BE}\approx 0.6V\$)
  • Base-Collector junction is reverse biased (\$V_{CB}>0\$)

Due to Base-Emitter junction being forward biased, there is an injection of electrons from the Emitter to the Base (\$I_{E_n}\$ in the image above), and the simultaneous injection of holes from the Base to the Emitter(\$I_{B1}=I_{E_p}\$ in the image above). Emitter region (\$n^{++}\$) is much more heavily doped than the Base region (\$p\$), therefore the current due to electrons injected into Base is much higher than the current due to holes injected into Emitter.

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.

Due to injection of huge amount of electrons from Emitter, electrons tend to diffuse through Base to Base-Collector reverse biased junction. Once an electron reaches there, it is swept across the Collector-Base depletion region and is injected into Collector thus contributing to Collector's current (\$I_C\$ in the image above).

Now, if all these electrons injected from Emitter could diffuse to the reverse biased Base-Collector junction without being subject to other effects - there were no importance at all to the width of the Base region. However, there is recombination going on in the Base.

In recombination process the injected electrons meet holes and "neutralize" each other. The injected electron is "lost" in this process and will not contribute to the current at Collector terminal. But wait, charge conservation requires that the hole which recombined with the injected electron will be supplied from somewhere, right? It turns out that the recombining holes are also supplied from the Base terminal (\$I_{B2}\$ in the image above) thus increasing Base's current and decreasing the Emitter-to-Base currents ratio (which represents transistor's current gain, remember?).

The above means that the more electrons recombine during diffusion through Base region, the lower the current gain of the transistor. It is up to manufacturer to minimize the recombination in order to provide a functional transistor.

There are many factors that affect recombination rates, but one of the most important ones is Base's width. It is evident that the wider the Base, the more time it will take to injected electron to diffuse through the Base, the higher the chance that it will meet a hole and recombine. Manufacturers tend to make BJTs with very short Base's.

So, why can't two PN diodes back to back function as a single NPN:

The above discussion explained why Base must be short. PN diodes (usually) don't have this short regions, therefore the recombination rate will be very high and the current gain will be approximately unity. What does this mean? It means that the current at "Emitter" terminal will be equal to the current at "Base" terminal, and the current at "Collector" will be zero:

schematic

simulate this circuit – Schematic created using CircuitLab

The diodes are functioning as standalone devices, not a single BJT!

Tricky question:

To various degrees of accuracy, many people can answer your initial question as I did. However, the more interesting question is this: if we make \$p\$ sides of both diodes very short, such that the sum of their widths will be no wider than Base region of NPN transistor, will the diodes function as a transistor?

This question is more difficult to answer because the straightforward answer of "no, the Base of BJT is very short" is not applicable anymore.

It turns out that this approach will not make two diodes similar in behavior to a single NPN transistor. The reason is that at metal contact of the diode, where metal and semiconductor are in touch, all the excessive electrons "recombine" with the "holes" supplied by the contact. It is not the usual recombination as metals don't have holes, but the fine distinction is not that important - once the electrons enter metal, no transistor functionality can be achieved.

The alternative way of comprehending the above point is to realize that Collector-Base diode is reverse biased, but still conducting high current. This mode of operation can not be achieved with standalone PN diodes which conduct a negligible currents under reverse bias. The reason for this restriction is the same - excess electrons from the P side of the forward biased diode can not be swept to the P side of the reverse biased diode through the metal wire in "BJT like diode configuration". Instead, they are swept to the power supply providing a voltage bias to the common terminal of the diodes.

There was a follow-up question which asked to provide a more rigorous reasoning for the above two paragraphs. The answer concerns metal-semiconductor interfaces and can be found here.

What the above means is that the discussion of the width of Base region is related to discussion of effectiveness of BJT transistors, and is completely irrelevant to the discussion of two back-to-back PN diodes as a substitute for a BJT.

Summary:

Two back-to-back PN diodes can't function as a single BJT because transistor's functionality requires semiconductor only Base region. Once a metal introduced in this path (which is what two back-to-back diodes represent) , no BJT functionality is possible.

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  • \$\begingroup\$ I'd thought of the difference as being one of geometry (by my understanding, BJTs are generally not symmetric, but have the emitter in the middle, surrounded by the collector which is in turn surrounded by the base), but it makes sense that transitor-ish behavior can't penetrate metal. \$\endgroup\$ – supercat Sep 24 '13 at 15:19
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    \$\begingroup\$ @vasiliy thanks for a great answer and putting me straight. It seems, as I read your explanation, that transistor action can occur on wider bases but, the current gain reduces because of recombination during base region diffusion and adding metal will completely stop transistor action. Thanks +1 \$\endgroup\$ – Andy aka Nov 17 '13 at 13:32
  • \$\begingroup\$ Thanks! The key insights for me were: (A) C-B junction is much weaker than E-B junction due to lighter doping of collector; (B) there are 2 types of current here: 1) driven by free electrons moving as they wish; 2) driven by bound electrons jumping from hole to hole. The free electrons don't have much problem passing the weaker C-B junction once they passed E-B. \$\endgroup\$ – akhmed Jul 11 '16 at 1:18
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No. Two back to back diodes is NOT a transistor. The special property that makes a PNP or NPN sandwich a transistor rather than just two diodes is that the base layer is very thin. In semiconductor physics terms, there aren't two separate depletions regions in the base. The depletion regions from the two junctions overlap in the base, which is necessary for the transistor to have its special properties.

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    \$\begingroup\$ Still doesn't explain why it doesn't work. \$\endgroup\$ – Curd Aug 8 '13 at 13:55
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    \$\begingroup\$ @Curd: It does, but apparently not at the level you want to know. I could answer with base current injecting carriers into the base region, but then you could ask why the carriers are needed, etc. We have to stop somewhere. A crude question gets a crude answer. \$\endgroup\$ – Olin Lathrop Aug 8 '13 at 14:01
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    \$\begingroup\$ I am just missing the keyword "diffusion", which is the basic principle in a BJT which also explains why it requires a thin base (see my answer). \$\endgroup\$ – Curd Aug 8 '13 at 14:07
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    \$\begingroup\$ In active mode and saturation mode the two depletion regions are separate. In Active EB junction is forward biased and the CB Junction is reverse biased, In saturation both junctions are forward biased and are at their minimal extent. \$\endgroup\$ – placeholder Aug 8 '13 at 14:31
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From Wikipedia

Transistors can be thought of as two diodes (P–N junctions) sharing a common region that minority carriers can move through. A PNP BJT will function like two diodes that share an N-type cathode region, and the NPN like two diodes sharing a P-type anode region. Connecting two diodes with wires will not make a transistor, since minority carriers will not be able to get from one P–N junction to the other through the wire.

Basically, the semiconductor needs to be connected directly.

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    \$\begingroup\$ Why would the minority charge carriers not be able to go from one PN junction to the other through the wire ? \$\endgroup\$ – user23564 Aug 8 '13 at 12:15
  • \$\begingroup\$ Because it is a long conductor instead of a small semi-conductor region. Olin's answer expands on this. \$\endgroup\$ – scld Aug 8 '13 at 12:56
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    \$\begingroup\$ The answer is correct, but Olin's answer is incorrect. The problem is not the length of the wire (which can be made very short, still not obtaining a BJT functionality), but the nature of metal as opposed to semiconductor. \$\endgroup\$ – Vasiliy Sep 24 '13 at 13:04
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It may be worth thinking about the equivalent question for vacuum tubes. Why can't two back-to-back diode tubes function as a triode? The answer is that for a triode to function properly, most of the electrons emitted by the cathode have to pass through the mesh of the grid in order to reach the anode. If you connected two diode tubes together and called the link between them a grid, or if you made the grid of a triode into a solid chunk of foil instead of a grid, then all the electrons would make it as far as the grid and stop there, draining out into the grid supply instead of being re-emitted to get to the anode. For correct operation of a triode there has to be an opportunity for the electrons' momentum to carry them right through the grid, driven by more than just the potential between the grid and anode.

The physical effects in play in a semiconductor transistor are different, but the fundamental idea that current must be able to bypass the wire that would otherwise suck it out in the middle remains the same.

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  • \$\begingroup\$ Aren't vacuum tubes more similar to JFETs than to bipolar transistors? In a tube, the gate acts as an anode, and if the grid voltage becomes more negative it will repel the electrons restricting their flow. There is no diode between anode and grid. The schematic here would be ................................ (A)--->|---(C)---|<---(G), the common point is the cathode rather than the grid. \$\endgroup\$ – Oskar Skog Mar 5 '17 at 18:00
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This is a very reduced version of the already accepted answer.

Metal has different properties than the semiconductor, so it will not unify the two N's into a single N. The Two diodes will be a P-N-metal-N-P component which is not an N-P-N component. (Vice versa for PNP.)

(If you slice the base of a transistor with a thin sheet of metal, it will stop working.)

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A BJT relies on the principle of diffusion (of minority charge carriers).

It works only if the thickness of the base is in the order of the diffusion length.

This can not be accomplished by connecting two discrete diodes.

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    \$\begingroup\$ You should mention the interplay between diffusion and minority carrier lifetime. \$\endgroup\$ – placeholder Aug 8 '13 at 14:17
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No, because to make a transistor needs only a thin layer between emitter and collector but if you connect 2 back to back diodes will provide a thick layer which will be difficult for electrons to penetrate

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Basically, one of the diodes will shut down due to the voltage difference in either base-emitter or collector (0.7 in any configuration). A closer approach will be a zener and two diodes but still won't function as a transistor nor anything useful. I'm terrible to explain, but the answer can be found in understanding how to eliminate voltage drop across a diode, something rarely found in books but extremely important. Now try to imagine a 0.7V battery in parallel with a diode connected to a signal then it will conduct starting from 0 and will collapse at 0 (not the typical -0.7). Well, there is more than that but I'm only trying to point you somewhere.

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