Claude Abraham
Member
Claude Abraham,
I have reasons for my beliefs other than what you think they are. I will concede that I did not explain thoroughly enough why Vbe is the cause of Ic. Although the equations are correct in showing that a Ib will always exist in a BJT operating in the active region, they do not show why voltage is the motivating factor. I will attempt to do that below. Let's not get into the AC characteristics of the transistor. The DC characteristics will suffice. I agree that Ib and Ic cannot exist independently in a BJT. I said so already, and the diode equation shows that.
I understand what you are saying, it just verifies that Ib and Vbe are related to each other. I already said as much. It still begs the question of whether Ib or Vbe controls Ic. I will explain that below. I know of no basic device that is a current amplifier except a magnetic amplifier.
Again you are referencing a model. As I said before, a model is good for explaining what a device does, but not how or why it does it.
I disagree. An ideal diode would not have any bulk resistance. It would still have a forward voltage due to the physics of the material. Ic is the majority part of Ie, so Ic does not control Ic, except in a model perhaps.
It flies in the face of the physics of the BJT. One cannot have zero Ib. I said that before. I also said that at very small values of Ib, the BJT behaves like a FET.
I never said that Ib can vanish. I said that Ib was inevitable, but not a controlling factor in Ic. I also said it was a indicator of what Ic would be, but did not control Ic. Ib is important for being a linear indicator of Ic.
Just to digress, I do not believe Ohm's law is V=IR. What you believe is Ohm's law, I call the resistance formula, and Ohm's law relates to the linearity of the material. In other words, it is a property of a material. My arguments to that are posted at the end of this thread. https://www.electro-tech-online.com/content/222-amp-hour-fallacy-ohm-s-law-noobies.html. You can contribute to it if you wish.
Of course the resistance formula can be written three different ways. But that does not mean that Ib or Vbe are both correct in what controls Ic. My further explanation follows.
In a NPN BJT, if the N-type material of the emitter is put in contact with a P-type material of the base, the electrons if the N-type material will migrate over into the P-type material material and vice versa. Eventually this will stop happening, because the N-type material will have positive ions along the boundary due to the absence of electrons that departed. These positive ions will keep the holes of the base from coming over. The same thing happens in the base in reverse. The electrons coming from the emitter neutralize the holes in the base and make negative ions which oppose further migration of the electrons from the emitter. Eventually a equilibrium is reached. These repulsive forces form a region along the NP boundary called the depletion layer. By applying a voltage Vbe, one can effectively make the boundary layer thinner and allow the charge carriers to move again. Or a opposite Vbe can increase the boundary layer and restrict the charge carriers. This is why I and others consider a BJT to be a voltage controlled device. Only a voltage on the base-emitter junction can do what needs to be done, that is, control the thickness of the depletion layer. Some electrons will get into the base circuit, the amount depending on the exponential rate of the diode equation. The Ic will also vary at a exponential rate. Therefore Ic will appear to be controlled by Ib in a linear manner. But a BJT is only mimicking a current amplifier. Nevertheless, this is a useful property that should always be used for design and analysis. But the operation of the transistor is dependent on the value of Vbe. Any bias arrangements or signal applied to the base will change Vbe, to a first approximation, by a few millivolts or microvolts. All the active region of the BJT is controlled exponentially by a Vbe of between zero and 1 volt. Therefore no one tries to control a BJT directly by a voltage source unless they are building a logarithmic amplifier. So, take it from there.
Ratch
But how does one establish a voltage on the emitter-base junction? You keep re-iterating that Vbe changes the depletion width. In order for Vbe to change, an external power source must do work moving charges through the b-e jcn. Take a PA system as an example.
The dc bias point is established in the microphone 1 stage preamp. The singer forms words, & her voice is amplified by a bjt. What gave rise to a change in Ic, namely "ic" the small signal component? It's not "vbe", nor is it "ib", nor "ie". They are all related & ic cannot exist w/o all 3 being present beforehand. However, the work required to change the depletion region is done by the singer. By moving lips & forming sounds, the acoustic pressure impinges on the mic diaphragm. The diaphragm vibrates, which is energy, or work being done.
This work results in a generation of current & voltage. The diaphragm motion generates the current/voltage that is the input to the mic preamp stage. The input resistance of the stage, including rbb' the base resistance, any emitter resistance present, & the depletion barrier are in series. Energy from the mic gives charge carriers the energy needed to transition from valence to conduction band. Since a p-n jcn is not ideal, there is a forward voltage drop. A portion of the energy is lost driving the b-e jcn. Likewise, some ib is needed to forward bias the jcn.
Thus ib, vbe, & ie, are all consequences of the work done by the singer. It is the singer doing work with her lungs & vocal cords, lips, etc. creating acoustic energy, that gives rise to ib, vbe, ie, as well as ic. The 3 quantities ib/vbe/ie are mutually inclusive. Neither 1 controls another. I've taken 1 solid state physics course from the phy dept as an undergrad. I've taken 4 such courses from the EE dept as a grad. I used highly esteemed texts, peer-reviewed, such as Kittel, Muller/Kamens, & Sze. Nowhere is it mentioned that Ib/Vbe/Ie have a pecking order, i.e. 1 does not control the other 2. Only contrarian non-peer-reviewed web sites like the 1 you referred to take such a position.
FWIW, you acknowledge that a bjt must be externally driven with current, not voltage. You agree that driving Vbe directly is thermally unstable. But when I refer to a bjt as current controlled, that is all I'm inferring. The current control view is only an external view, looking at the device from a circuit perspective. Current control was never claimed to be a microscopic atomic viewpoint.
At the atomic level, only quantum mechanics, QM, can explain device behavior. The ultimate model is Kronig-Penney. The equation is that of Schroedinger. My current control model is just used to externally drive the device. I never said it was valid at atomic level physics. As long as we understand that, all is well. It is also important to state that QM is the only valid FET model when viewing atomic action. The FET voltage control model is a good external model for driving the FET device. It cannot explain semiconductor physics at atomic scale.
Also, FWIW, in a log amp, we do not control voltage, but current. We measure the Vbe with a specific value of current. I spent years working w/ log amps, & I have a patent. Look up US no. 5,670,775. It uses a diode as the log element. Don't be angry with me, but how much formal education have you in semiconductor physics? You keep saying that this controls that, yet you cite no verifiable laws, or peer-reviewed texts. You earlier stated that I & others present valid facts, but err when drawing conclusions. But that is what you are doing, except that not all of your info is factually right. BR.
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