Why building a BJT transistor out of two diodes is impossible?

In summary, the idea of using two diodes to create a BJT transistor is not electrically equivalent. This is because the BJT requires a thin base region for carriers to travel through, while in two separate diodes, this is not possible. Additionally, the process of minority carrier injection is necessary for transistor action, which is not present in two diodes. The BJT is also a voltage-controlled amplifier, while a wire is a current amplifier. Ultimately, the base region must be directly connected to the collector region and be thinner than the minority carrier diffusion length for a transistor to function properly.
  • #1
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i had a thought to stick two diodes together such that the electronic setup was similar to that of a BJT transistor. however i have been told by multiple teachers/professors that the two were not equivalent. I have drawn a picture to show you what i mean. the top two pictures are just the circuit setups. the bottom two pictures are the physical electronic setup that is being represented.
Please explain why these are not electrically equivalent.

http://i.imgur.com/6ssocN3.png

so basically my rationale was, that since a conductor is supposed to carry charge, ie, where ever there is an excess of charge, there becomes a potential difference and thereby allows the charge to travel through the conductor. so whenever one of the P material has an excess or a defficiency of charge, it will get transferred over to the other P type material in the other diode. and.. it gets transferred over through a conductor which of course can be tapped, thereby able to be turned into a BJT transistor setup.
 
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  • #2
The basis has to be very thin in a transistor so that the charge carriers which enter from the collector into it have a chance to leave it through the emitter before recombining in the basis.
With two separate diodes, this isn't possible.
 
  • #3
Correct answer, except that charge carriers are actually injected from the emitter and diffuse through the thin base to the collector. (John Bardeen, coinventor of the transistor, purposefully chose names for these regions that are descriptive). The process is formally called minority carrier injection because the carriers injected from the emitter (electrons for an NPN transistor) form a minority in the base (which is P type for an NPN).

The minority carriers start to recombine in the base but many diffuse through to the collector if the base is thin enough. This is why two separate pn junctions can't display transistor action.
 
  • #4
What a BJT does:

There's a bias between the emitter and collector. But current can't flow because the base presents an energy barrier to carriers (e.g. for NPN, the P region is higher than N in an energy diagram). But if you were to inject current into the base, this charge would build up and applies a potential that lowers the barrier. Carriers rush in from the emitter to restore local charge neutrality, but most of them shoot through the base without finding an opposite carrier to recombine with in time. This is why a small I_b causes a large I_c and I_e.

In contrast, the wire is a metal. It takes a long time to move through a metal because of scattering and the sheer number of carriers means it's very easy to find a recombination partner. If you shoot current into the wire in between the two bases, it just goes out the two diodes. No amplification is taking place because that's a manifestation of base transit times being shorter than recombination time. Also the currents I_b and I_c are in opposite directions, instead of the same as in a BJT (diodes only conduct one way and these two are facing opposite directions). There's no way to make a BJT like this.
 
  • #5
erst said:
What a BJT does:

There's a bias between the emitter and collector. But current can't flow because the base presents an energy barrier to carriers (e.g. for NPN, the P region is higher than N in an energy diagram). But if you were to inject current into the base, this charge would build up and applies a potential that lowers the barrier. Carriers rush in from the emitter to restore local charge neutrality, but most of them shoot through the base without finding an opposite carrier to recombine with in time. This is why a small I_b causes a large I_c and I_e.

In contrast, the wire is a metal. It takes a long time to move through a metal because of scattering and the sheer number of carriers means it's very easy to find a recombination partner. If you shoot current into the wire in between the two bases, it just goes out the two diodes. No amplification is taking place because that's a manifestation of base transit times being shorter than recombination time. Also the currents I_b and I_c are in opposite directions, instead of the same as in a BJT (diodes only conduct one way and these two are facing opposite directions). There's no way to make a BJT like this.
There are a few misconceptions here. First, a bias is an externally applied potential, not an internal one. Second, the EB barrier is lowered by the applied bias, not by a charge build up. Third, a large Ic is not exactly caused by a small Ib. Instead, a large current is sourced from the emitter. You are correct that most of this current flows to the collector, so that only a small base current is needed to make up for charge carriers lost to recombination in the base. It is said that the BJT is a current amplifier, but internally it is in fact a voltage-controlled amplifier.

Despite scattering, net current flow through a metal is nearly instantaneous (it propagates at close to the vacuum speed of light.) Furthermore, there is no recombination to speak of in a metal conductor, so the description in the second paragraph misses the mark. The reason that two discrete diodes don't make a transistor is because the base region must be abutted to the collector region and must be thin compared to the minority carrier diffusion length.
 
  • #6
marcusl -- The current picture is fully equivalent to the voltage picture. See e.g. Streetman Ch. 7. It's meaningless to say I_c is not caused by I_b; local (steady-state) neutrality demands it, so it's just as causal as a bias-based description. It's helpful to think of what would happen if a current was injected into the base and flood of carriers *didn't* come in from the emitter. The carrier build up would change the chemical potential, i.e. a bias has been applied. Which way cause-and-effect goes is not meaningful since chemical potential is a bookkeeping tool.

Metals do have recombination and it is the transit time of individual carriers that matters, not the signal velocity. There's a thermal distribution about the Fermi energy with constant scattering in and out of states above and below it. That's simply generation and recombination in semiconductor terminology; the term is not used because the incredibly high scattering rate (relaxation on the order of ~10 fs) makes it nearly meaningless. That scattering rate is also precisely why it wouldn't work, not the absence of a gap. A short enough piece of high relaxation time metal will preserve and transmit a hole or electron current (if the Fermi level lies within the contacting materials' gaps). It's not the material category (metal v. semiconductor), but lifetime v. length scale that prevents BJT action.
 
  • #7
guys guys! would if i cut into the cathode part of both diodes far enough such that the total amount of P-type material is small enough in comparison to the N s.t the charge carriers can pass right through the base to the collector, and only a few electrons would get carried out through the base?

like so!

http://i.imgur.com/eKCvl2n.png...well...except... just imagine that the P-type was cut into a lot more
 
  • #8
Well that's the idea. However sticking two pieces against each other creates huge impediments (scattering centers and surface traps) to current flow. There's a reason that classical transistors are made of extremely pure and nearly perfect crystals of semiconductor material.
 
  • #9
marcusl said:
Well that's the idea. However sticking two pieces against each other creates huge impediments (scattering centers and surface traps) to current flow. There's a reason that classical transistors are made of extremely pure and nearly perfect crystals of semiconductor material.

i couldn't get a simple definition/description of what S parameter is; what is the S-parameter?
 
  • #10
You mean microwave S-parameters? Please start a new thread--this is a new topic.
 

1. Why can't a BJT transistor be built using just two diodes?

A BJT (Bipolar Junction Transistor) is a three-terminal device, which means it has three layers or regions of different semiconductor materials. Two diodes, on the other hand, only have two layers. Therefore, it is physically impossible to create a BJT using just two diodes.

2. Can't the two diodes be connected together to create a third layer?

No, connecting two diodes together does not create a third layer. Diodes are made of a single p-n junction, while a BJT requires two p-n junctions that are joined together in a specific way to create the three layers. Therefore, simply connecting two diodes will not result in a functional BJT transistor.

3. Why can't a BJT be made by using two different types of diodes?

While it is possible to create a p-n junction using two different types of diodes, it still does not create the necessary three layers for a BJT. The layers in a BJT must be made of specific semiconductor materials in order to function properly, and using different types of diodes will not achieve this.

4. Are there any other ways to create a BJT without using three layers?

No, the three layers are essential for the operation of a BJT transistor. It is not possible to create a BJT using only two layers or using a different type of device.

5. Why is it important to have three layers in a BJT transistor?

The three layers in a BJT serve different purposes. The two outer layers, called the emitter and collector, are heavily doped to allow for efficient flow of charge carriers. The inner layer, called the base, is lightly doped to create a depletion region that controls the flow of charge carriers from the emitter to the collector. Without all three layers, the BJT would not be able to function as a switch or amplifier.

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