Why can a transistor be considered to be made up of diodes?

Question

In my electronics book, it says that a transistor can be considered to be made up of either two back - to - back or two front to front diodes, but I fail to see how current can flow through a transistor from collector to emitter then.

In a circuit with two front to front diodes no current would be able to flow, so how come it can in a transistor? Why is the base current not included in the "output" i.e the collector or emitter current?

EDIT: I just realised the book is talking about digital signal to the base is not included. It's not talking about the current.

My textbook said that it is a common mistake students make. That the base current is not included in the collector or emitter currents — Jan F. S, Apr 03 '23 at 22:50
You'll have to ask them why they make the mistake, then. Probably because it's small relative to the emitter current, but it's still significant. — Finbarr, Apr 03 '23 at 22:54
a transistor is only two diodes if you're using it in a really weird way. Most of the time if you wanted two diodes you'd use two diodes. It *is possible* to bias a transistor so it acts as two diodes, but... why? — user253751, Apr 03 '23 at 23:02
@JanF.S Your textbook may not be wrong, but if your textbook says that students commonly forget base current, then that's what it says. We don't know the percentage of people who forget base current, as if you do that often enough, you fail to pass the exams. — Justme, Apr 03 '23 at 23:02
@Justme I'm asking for clarification as to why the base current isn't considered in the collector or emitter outputs like my book says it's not. If you don't know why that's fine, and while I appreciate people taking the time to comment, it's not very helpful to write something so irrelevant. — Jan F. S, Apr 03 '23 at 23:10
@JanF.S Could you post an image of the passage from your text? Thanks. — RussellH, Apr 03 '23 at 23:15
@JanF.S I don't know what your book says but typically all books say Ie = Ic + Ib but sometimes you don't need Ib and can approximate that Ie = Ic. It depends if you are in the first chapters or somewhere else in the book, the book likely explains why it has the equations it does, or then it's not a good book. — Justme, Apr 03 '23 at 23:16
@Justme I'm looking through my notes and my teacher did write Ie = Ib + Ic. So I dont know what my book means then... — Jan F. S, Apr 03 '23 at 23:28
@RussellH Nevermind guys, I just realised my mistake. The book is about digital signals, and that is what is meant by that the base signal is not part of the collector or emitter signals. Not that the current Ib is not included. My bad. — Jan F. S, Apr 03 '23 at 23:32
Duplicate? https://electronics.stackexchange.com/questions/78366/why-cant-two-series-connected-diodes-act-as-a-bjt — Sredni Vashtar, Apr 04 '23 at 01:23
Perhaps the textbook means that students think the base current goes through \$R_c\$, or like, they calculator the emitter current using a usual formula, and then add the base current onto whatever the formula says, when it's actually already included in the formula. — user253751, Apr 04 '23 at 11:17

score 26 · Accepted Answer · answered Apr 03 '23 at 23:11

The BJT works because the base is all one piece, thin, and connected to the emitter and collector.

When you forward-bias the emitter-base junction, carriers flow from the emitter into the base, as if they were all going to get recombined in the base. But -- because the base is thin -- they're now in what would be the base-collector junction's depletion region. In a well-designed transistor, this means that most of them end up passing through the base into the collector, and a fraction end up contributing to base current.

If you just wired two diodes back to back this would not happen, because you don't have that one-piece, thin base.

It's worth pointing out that one failure mode from "secondary breakdown", a thermal runaway condition in power transistors, is that semiconductor areas can get destroyed to the degree that two independent diodes remain. This condition can be quite misleading. It also tends to blow the less powerful driver transistors that have to bear the full load current then. — user107063, Apr 05 '23 at 15:31

score 8 · Answer 2 · answered Apr 03 '23 at 23:12

Your textbook is absolutely correct.

A BJT can be thought as just two PN junctions slabbed together, and a diode is just one PN junction.

When looked externally as a black box and measured with say a multimeter in diode mode, you can observe that a BJT transistor looks like there are two diodes connected together.

Now, with that arrangement, and given with two black boxes, one with a BJT in it and one with just two diodes in it, you likely can't tell apart with the multimeter which one is a BJT and which one just has two diodes.

So while a BJT can be considered as just two diodes, it is obviously only an observation and of course a real BJT will behave differently in a circuit than just two diodes.

So it is not an useful model for what a BJT is or what it does in a circuit.

However, if you need to determine from a bunch of transistors if they are OK or broken, it is a very useful model, as you can directly discard the ones that don't seem to have two working PN junctions, if you only have a multimeter with diode measurement.

"you likely can't tell apart with the multimeter which one is a BJT and which one just has two diodes." Does nobody teach people how to test transistors with a multimeter these days any more? Resistance mode. NPN: plus on collector, minus on emitter, thumb bridging collector and basis. Result should be considerably less resistance than thumb alone. For PNP, swap plus and minus from the multimeter. — user107063, Apr 05 '23 at 15:23
@user107063 I did not say it is impossible and for this question it is irrelevant. The point was not how tell them apart but how a transistor will look like two diodes if you try it with a multimeter in diode test mode. Sure, with correct technique you can determine which one is base, collector and emitter by just using a multimeter. — Justme, Apr 05 '23 at 15:28

Jasen Слава Україні · Answer 3 · 2023-04-03T23:15:30.737

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how current can flow through a transistor from collector to emitter

Keep reading, that bit will be explained later, just know that in a two terminal measurement that the transistor will appear to be made of diodes.

edited Apr 03 '23 at 23:15

answered Apr 03 '23 at 23:13

Jasen Слава Україні

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Which property is very handy if you have a junk box with unknown transistors in it, and you want to identify the pins (if you do this, note that the base-emitter junction has a lower voltage drop from the base-collector junction, so even if it's an oddball pinout you can still get the pins in the right places in your circuit). – TimWescott Apr 03 '23 at 23:15
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@TimWescott And if you only detect a diode between one pair of pins, and an open between the other two pairs, then you have a MOSFET, not a BJT. And if you measure diodes like it would be a BJT, but then the "collector" and "emitter" measure a short between them, it's a JFET. – Hearth Apr 04 '23 at 00:48
@Hearth "And if you only detect a diode between one pair of pins, and an open between the other two pairs, then you have a MOSFET": unless you take a lot of antistatic care while handling, most likely a blown MOSFET. It takes rather little charge to load the base capitance beyond its tolerance level. – user107063 Apr 05 '23 at 15:36

score 5 · Answer 4 · answered Apr 04 '23 at 15:44

The two "diodes" that make up a transistor interact in a manner that could be viewed as (very) vaguely analogous to the coils in a transformer. If one measures the current/voltage response of each coil of a transformer separately, while the other end isn't attached to anything, each coil would behave as an inductor. What makes transformers useful, however, is the interaction of the two coils.

Likewise, what makes a bipolar transistor useful is the interaction of the two diodes. If one thinks of a diode as being like a flap valve, which will be pushed open when voltage is applied across it one way and pushed shut when voltage is applied across it the other way, one can think of the two diodes in a transistor as having a linkage between their flap valves in the parts of the base that are closer to both the collector and emitter than to the metal connection. In an NPN transistor, when electrons flow from the emitter to the base, that pulls open the valve between them, which in turn pulls open the valve between the base and the collector, allowing electrons to flow there as well.

If one were to connect two separate PN diodes together, joining the P-doped regions, there wouldn't be any portion of any P-doped region that was closer to both N-doped regions than to the metal attachment point, and thus no area where the "valves" created by PN junctions would be tied together. Expecting two diodes to behave as a transistor would be somewhat like expecting a two separate inductors to behave as a transformer.

score 3 · Answer 5 · answered Apr 03 '23 at 22:46

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The typical diagram used to show a BJT betrays the actual physical shape. The emitter is much more heavily doped than the other regions, not equal to the doping of the collector, and the base is very thin. Because of this, the behavior of the two PN junctions (the two "diodes") are dramatically different. In this way, a BJT is not symmetrical in the way a MOSFET is.

answered Apr 03 '23 at 22:46

Aaron Linnell

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AFAIK, it's the thin base that is the key. The geometry and doping of the collector and the emitter can help a lot, but if you run a transistor "backwards" (hook up the collector where the emitter should be, and visa-versa), there's still some observable current gain. This was actually used to advantage in TTL logic, to help turn off the input transistors quicker. – TimWescott Apr 03 '23 at 23:13
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@TimWescott the thin base is important, but so is the doping level. Generally, the base doping level is orders of magnitude lower than the emitter doping level. As a result, the minority carriers in the base (arriving from the emitter) vastly outnumber the majority carriers, and so recombination is "slow" according to the scale of things. – Math Keeps Me Busy Apr 03 '23 at 23:28

score 0 · Answer 6 · answered Apr 04 '23 at 21:12

Book is correct, but incomplete in explanation. A BJT transistor can be viewed as made of two diodes, correct, but the book misses the important part that these diodes have to be arranged in a specific manner (physically fused together) that cannot be achieved in any way by just connecting the external leads of those two diodes.

All BJT transistors are made of two diodes, but not all two-diode arrangements are transistors.

score 0 · Answer 7 · answered Apr 05 '23 at 15:48

Another way of thinking about things. Both are created by doping electrically neutral material with either Negative of Positive charges (N/P type). This configuration allows for electrons to flow in useful ways. Do you notice anything interesting about the following graphics?

Here's a Diode:

Here's a BJT:

The NPN BJT has the same PN junction that the diode has as well as another junction going to the collector.

score -1 · Answer 8 · answered Apr 05 '23 at 03:03

Mate, a transistor is like two diodes that are connected together. It has three terminals: emitter, base & collector. Emitter & collector are like the anode & cathode part of two diodes that are connected together made from different type of semiconductor material - one is doped with impurities to make it a p-type material, and other is doped to make it an n-type material. When these two regions are connected, they form a p-n junction just like in a diode.

The base is middle part of the transistor, & made from different semiconductor material either p-type or n-type. Its not like a diode, but its kind of like resistors where can control flow of current through the p-n junction between the emitter & collector.

So when apply voltage to the base of transistor, it can control flow of current between emitter & collector, same like turning a tap on/off, arigato goza

Why can a transistor be considered to be made up of diodes?

8 Answers8