crutschow,
A low input impedance makes the device amenable to a current source. It does not change whether it is voltage or current controlled. Current controls the magnetic amp, and voltage controls the BJT.
So why does a BJT operate better and better when the β becomes higher and higher and Ib become less and less? Ultimately, when the β becomes super high, the base current becomes insignificant and the BJT behaves like a FET. It does so anyway for very low base currents. You still have not explained the relationship of the equation that relates the Vbe to Ic.
I never said that folks drive a BJT with a voltage source in the active region. That would be folly due to its very nonlinear exponential voltage-current characteristics. Everyone adds external circuitry to linearize the output. That does not take away the fact, as demonstrated by the Vbe to Ic equation, that the basic BJT is a voltage to current responding unit, which mimics a current amplifier.
That is not the reason a BJT is a voltage controlled device. It is because voltage, not current, controls the collector current, as demonstrated by the aforementioned equation. By the way, there is no significant voltage involved with superconductor currents.
Perhaps, but I can be convincing. But can you or anyone else defend your position?
Ratch
First of all, Ib is NOT a "waste" current. Without Ib & Ie both, Vbe can never be established. How does Vbe go from 0.00V to 0.65V, for example? The answer is that charge must be transported through the base-emitter region. Upon crossing the junction, the carriers which were formerly majority type become minority type since the material is different. Holes from the base are majority carriers for an npn bjt, & electrons from the emitter are majority in the emitter. But they cross the junction & holes enter the emitter, while electrons enter the base, & they are now minority carriers.
The associated electric field due to the stored minority carriers tends to oppose forward current in the b-e junction. When unbiased, thermal energy generates electron-hole pairs & they cross the junction resulting in an electric field which opposes forward current. The integral of said field over distance is labeled "thermal voltage" aka "built in potential". At room temp it is roughly 25.7mV.
When forward biased, many more charges flow & the junction potential is around 0.65V, give or take w/ temp, & current value. In order for Vbe to be non-zero, a base current & emitter current must first take place. Vbe comes after Ib & Ie. This component of base current is called
displacement current. It is fundamental & a bjt cannot operate w/o it. It is like the gate current in a FET. The FET gate current is absolutely indispensable & fundamental for FET operation.
With bjt & FET, a transition frequency, ft, is defined on every data sheet I've seen. This ft parameter is by definition the frequency where the current gain has fallen to unity (0 dB). A device with current gain less than 1 is not very useful as an amplifying device, so ft is stated for that reason. Displacement component of base current is all important & a bjt cannot operate w/o it. Ib is not a residue or "waste" current.
The holes injected from base to emitter do not contribute to collector current. But we can minimize this injection component of current by doping the base much lighter than the emitter. Also, the transport component of base current must be discussed. Electrons emitted from the emitter towards the base diffuse & are drawn into the collector by its local E field. A few electrons recombine in the base never reaching the collector. So the same number of electrons must exit the base to preserve charge neutrality. Otherwise the base region would build up an unbalance of charge & bjt action ceases. But the transport component of Ib is generally much less than the injection component. A super thin base region minimizes said transport component.
So the transport component of Ib is very small, but fundamental to operation, as is the displacement part of Ib. W/o these base current components, there is no bjt action. Vbe is also indispensable as well. But Vbe can also be viewed as a "waste" quantity.
Ideally, a forward biased p-n junction, such as a b-e jcn, carries all current & zero voltage. The notion that a bjt should have positive non-zero Vbe with zero Ib is ludicrous. Such a device is not even BI-polar, but unipolar such as a FET.
Anyone who has ever worked with amplifier bjt power output stages has learned about cross-over distortion 1st hand. In the +/- 0.7V signal region, the device has to cross over. Many papers, patents, articles, books, networks, etc. have been coined to address this issue. There is the class A, AB, B, etc. approach. The objective comes down to 1 thing.
If the Vbe was what it ideally should be, namely zero, all this trouble would not exist. I've designed many networks where the 0.7V Vbe drop was an obnoxious limitation, & wished that a germanium device with 0.3V drop was available. Vbe can be just as much a nuisance as is Ib. Also, with op amps, there is an output error due to Ib mismatch, and one due to Vbe mismatch. If Ib was zero, there is no offset current error. But likewise, if Vbe is zero, there is no voltage offset error.
This notion that Ib is just a "waste" product, is propagated by wannabes, hackers, semi-literates. pretenders, & know-nothings. In the ideal realm, Ib & Vbe both vanish, leaving Ie as the control parameter for Ic. But in the real world, we are stuck with Ib & Vbe whether we like them or not. Still, it is Ie that controls Ic.
The Ebers-Moll equations, published in Dec 1954, describe a bjt as 2 current controlled current sources. Although you can derive Ic as a function of Vbe, it is also true that Ic is related to Ib, as well as Ie.
1) Ic = beta*Ib
2) Ic = alpha*Ies*(exp((Vbe/Vt) - 1)
3) Ic = alpha*Ie.
In equation 1), forcing a specific value of Ib results in Ic per 1). But this approach is avoided most of the time due to beta dependency. It is hard to predict beta with differing speciman & temp. In 2), Ies varies with speciman, hugely with temp, & a bjt driven w/ a voltage source across Vbe is thermally unstable. A Vbe connected to a source, results in Ic, some power, a temp rise, then Ies increases due to elevated temp, Ic increases, temp increases, etc. Thermal runaway means that you can never "voltage control" a bjt.
In eqn 3), alpha is very stable in value. Controlling Ie results in a very predictable Ic. It is thermally stable & predictable. Ic is ultimately controlled by Ie, not Vbe.
To see this, take a bjt w/ a very thick base region. A Vbe of 0.7V is set up, & Ie = 10 mA, which also equals Ib. Because the base is very thick, alpha=0, & all electrons emitted are recombined in the base, so Ic = 0. "Transistor action" does not occur until electrons emitted from the emitter are "collected" by the collector. Although Vbe=0.7V, there is no Ic.
What controls the number of electrons in the collector is the number emitted from the emitter. If 1010 electrons are emitted in 1.0 picosecond, then the Ic value can never exceed 1010 e-/psec, regardless of Vbe. The emitter emission literally controls the collector collection.
A bjt that is "voltage driven" is thermally unstable, & unpredictable. A bjt that is "base current driven" is thermally stable, but beta dependent. Still we use base current control for saturated bjt switches. By forcing a base current larger than Ic/beta_min, we assure saturation, & beta dependency is mitigated. A bjt that is "emitter current driven" is thermally stable & predictable. The value of Ic is influenced by temp, Ib, Vbe, doping, illumination, Vbc (Early effect), but Ie is the parameter that chiefly controls Ic.
Again, the number of electrons per picosecond "collected" by the collector is proportional to the number "emitted" by the emitter. The emitter & collector leads are so named for a very good reason. Questions/comments are welcome.
Claude