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But I find it to be absolutely incorrect. How can a BJT transistor even work without a base? Shouldn't a transistor without base be just a semiconductor (PP, NN)? Does there exist some special NASA-army-grade experimental BJT transistor without a base?

So who is right, me or the teacher?

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    \$\begingroup\$ A base connection is often omitted but it still has a base. \$\endgroup\$ – Andy aka May 10 '17 at 12:38
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    \$\begingroup\$ Did he say "base" or "base connection"? \$\endgroup\$ – CL. May 10 '17 at 12:39
  • \$\begingroup\$ he just mentioned base \$\endgroup\$ – qwerty12456 May 10 '17 at 12:46
  • \$\begingroup\$ You are right. 3-pin phototransistors have a base wire, but PTs in 2-pin packages don't. Yet all PTs have a base region. Go find a datasheet for 3-pin package TIL99, or this NTE replacement weisd.com/store2/NTE3032.pdf \$\endgroup\$ – wbeaty May 10 '17 at 18:52
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Clearly the source of confusion here is what base means: you assume it's "base region", and your teacher seems to assume "base terminal".

What defines a BJT transistor is the NPN or PNP structure. How many of these regions are exposed through terminals is irrelevant: most transistors will have three, but phototransistors may have only two, and parasitic transistor structures may not be exposed at all, but they still behave similarly and are considered to belong to the same device class.

Phototransistors are a bit special in that they have no base current in the strict sense, as carriers are generated in the base itself. But since the structure still behaves quite similarly to a regular BJT, this fact is usually ignored and photodiode current generated by light is considered to be base current.

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I think you and your teacher are both right - you are just not using the same definitions and thus your wording disagrees.

From your description:

Shouldn't be transistor without base be just a semiconductor (PP,NN)?

You define the "base" to be the semiconductor in the middle of a normal BJT - the P-doped region in a NPN or the N-doped region of an PNP. In this regard you are right: A phototransistor still has this PNP or NPN structure.

Your teacher might define the base differently - the terminal you put current through to achieve a collector-emitter current. When you look at it this way, he is also right - most phototransistors (I know of none myself that break this rule, apart from some optocouplers) have no "base" lead. The only way to trigger a current flow is through photons hitting the base region, and causing the release of electrons by doing so, turning the device on.

Both make sense, although I would argue that you need to watch out when thinking that "no base" means "pp or nn" semiconductor. Many complex semiconductor structures exist that have many regions, yet have few leads. Look at Triacs or IGBTs!

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    \$\begingroup\$ A "transistor" seems to be defined as a 3-terminal semiconductor, including bipolar,JFET,MOSfet devices. But I'm siding with Joren - a phototransistor falls between the cracks of this definition - the base connection is not required, but a "base" doping region is required. \$\endgroup\$ – glen_geek May 10 '17 at 16:48
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    \$\begingroup\$ Used to see lots of phototransistors that had a gold dot on the bottom where the base pin would be in a normal transistor. I always figured the cut them off in the factory. \$\endgroup\$ – JRE May 10 '17 at 17:20
  • \$\begingroup\$ His teacher is simply wrong. TIL-99 has a base wire. "PHOTOTRANSISTORS IN 2-PIN PACKAGES HAVE NO BASE TERMINAL." I doubt this is what his teacher was saying. \$\endgroup\$ – wbeaty May 10 '17 at 17:58
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The base connection is often not used and would thus serve mostly to pick up noise.

You can find optocouplers (4-pin) which have no base connection and others (eg. 4N35) which bring the base connection out (where it is sometimes used).

enter image description here

Similarly, individual phototransistors are often packaged in an LED-like packages with lens and 2-pin leadframe.

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  • \$\begingroup\$ "where it is sometimes used"... what would be the purpose then, to override or offset the optic signal? \$\endgroup\$ – dlatikay May 11 '17 at 15:01
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    \$\begingroup\$ @dlatikay You can use it as an input to 'or' with the optical signal. It reduces the sensitivity (not usually a good thing) and speeds up the operation (usually desirable) if you add a (usually relatively high value) resistor from B to E. If you use just C and B you have a photodiode, which could be very fast compared to the transistor- use an external transimpedance amplifier. Lots of possibilities. \$\endgroup\$ – Spehro Pefhany May 11 '17 at 15:43
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Your teacher right. BJT photo-transistors rarely have base connection.

It's the light hitting the base region that controls the current flow from collector to emitter. The more light, the larger the current flow. That's what makes it a photo-transistor and not just a transistor.

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    \$\begingroup\$ No base connection, but base is present :) \$\endgroup\$ – rackandboneman May 10 '17 at 13:56
  • \$\begingroup\$ @rackandboneman, quite and as per my answer... \$\endgroup\$ – TonyM May 10 '17 at 14:23
  • \$\begingroup\$ If the teacher didn't say "rarely," that makes all the difference. And only in recent decades has it become rare. Years ago 3-pin PTs were common: see the old texas inst. TIL-99 \$\endgroup\$ – wbeaty May 10 '17 at 18:51
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    \$\begingroup\$ @wbeaty, I'm lost. I said 'rarely' as above. I spoke about the base connection and the base region. Yes, I remember them clearly from years ago but this is a contemporary question. \$\endgroup\$ – TonyM May 10 '17 at 19:02
  • \$\begingroup\$ @TonyM The OP and their teacher didn't say "rarely." Their question implies "never." Their teacher may actually be wrong, or at least was accidentally misinforming their class. OP may want to get it clarified in a public discussion, since others in the class may be equally confused. \$\endgroup\$ – wbeaty May 10 '17 at 19:21
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Base is already present. Connection to base is present too, but it works not from electricity - from the light. No matter how the energy comes into transistor, but it MUST come someway to make it work, so you can think of base connection as the wireless to the Sun )

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  • \$\begingroup\$ Calling a photosensitive region a "connection" makes little sense since there is no corresponding current. \$\endgroup\$ – Dmitry Grigoryev May 11 '17 at 11:22
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I use the base pin on a phototransistor to handle DC currents from the sun, or 60Hz currents from incandescent lighting.

schematic

simulate this circuit – Schematic created using CircuitLab

By nulling out the DC error, you can use high value collector resistors thus achieve high gain.

An alternative is that current source I drew in parallel with the collector resistor, to be activated by Vout lower than e.g. VDD/2

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