More than that - all measurements, experiments, observations proof this (Early effect, -2mV/K, Re-feedback, diff. amplifiers with tanh-transfer, class-B form of cross-over distortions, current mirror, translinear loops,...). In particular, the design of BJT stages is based on voltage control. Everybody knows that he must provide a suitable B-E voltage to allow the desired current Ic.
I wonder how current injection into the emitter (without suitable B-E voltage) could allow transistor operation (as preferred by Claude - however, without any example circuit).
The question was about a dc motor (voltage controlled?). It was my attempt to understand Claude`s way of thinking. Perhaps he didn`t recognize - but I gave an answer in my post#77.
However, his answer was (a) "I didn't answer so as not to drift off topic" and (b) that I have "a futile position" - whatever that means.
I've answered all of those. But to have it in a neat single post, concise form, here it is.
Early effect is not a "Vbe" issue, but rather "Vbc", or "Vce". Early effect has a minor influence on Ic wrt "Vbc". But Early effect has been acknowledged since it was published in 1952. The fact that Vce has a slight influence on Ic has never been debated. That is not the main control quantity, Ie is. Also, the slope of the Ic vs. Vce curves is often measured with Ib as the parameter held constant. Vbc/Vce when increased results in the depletion layer of the reverse biased b-c junction larger. It encroaches on the base region, reducing its volume. As a result beta increases due to less injection hole current from base to emitter. But these curves are generated with Ib constant, so emitter current increases due to increased beta. I want to emphasize that amp stages are usually designed to fix the emitter current, not base current.
When Early effect takes place, beta can increase by 10 or even 20%. But if a bias network fixes Ie q-point, then the collector current Ic is given by Ic = alpha*Ie. Remembering that alpha = beta/(beta+1), if beta is 100, alpha is 100/101, or 0.9901. Then due to Early effect, beta increases to 120, so that the new alpha value is 120/121 or 0.9917. So Early effect increased alpha from 0.9901 to 0.9917, an increase of only 0.165%. If a bias network is well designed, fixing Ie at a stable q-point, Early effect, die to Vce/Vbc, has very minimal influence on Ic.
Still, it is correct to say that the collector-base **voltage** does exert a slight amount of "control" over Ic. Slight, but nonetheless, it is safe to say that Ic is to a small extent, influenced by Vbc/Vce.
The 2 mV/K issue, how does that prove anything re control. A bjt is biased with a network fixing Ie, not Vbe. For a given Ie, we know that the Vbe value changes by 2 mV/K temp change. If a bjt was controlled by fixing Vbe (voltage controlled), the temp coefficient would be in **mA/K**.
Re feedback is one I already discussed. A q point is established, but the bjt is suddenly heated up. What happens? Silicon when heated becomes more conductive due to increased carrier concentration due to increased thermal energy in the form of lattice vibrations. As soon as current increases, the Re incurs more collisions resulting in ionization and an E field opposed to signal source. This reduces Ie as well as Ib, and eventually Vbe is lowered. But the net result is an increase in Ie and Ic, with a decrease in Vbe. Less Vbe is the result of the increased ionization of the resistor. THis reduced Vbe does not bring Ic down as you claim. Measure it in the lab, Ic increases at elevated temp, but Vbe goes down. Also we can look at it as the dc source voltage at the base Vbb is fixed. Ve is Vbb - Vbe. Ie is (Vbb-Vbe)/Re = Ve/Re. The drop of Vbe will result in a slightly larger Ve, and larger Ie. Ultimately Ic goes UP due to Ie going up. Vbe going down does not bring Ic down. What happens is that a nw equilibrium is reached. Ic can increase even though Vbe decreases because Ies went up a lot, while Vt went up as well.
But please remember that this Re feedback works only because the base terminal is forced to a constant voltage Vbb. If the base were driven by a current source, a bjt collector, or another CCS, the Re does not provide this feedback. Taks a diff pair 1st stage in an op amp, with a common emitter 2nd stage. The diff pair collector of one bjt inputs to the 2nd stage base of the common emitter. Do a sim, or set up a lab test, you will find that Re does not provide said feedback. When beta varies, or temp, the change in Ic is NOT cancelled. This type of topology is beta dependent, which is why op amps open loop gain varies, due to 2nd stage beta variation. Re does not mitigate that.
Dif amp tanh relation proves what? That an amp built from 1 or more bjt's has voltage gain as well as current gain. I have the text by Schaub and Tilling I will post tonight showing the current gain transfer function of a diff pair. If the stage is used to serve as an amp front end, popular with op amps and discrete audio power amps, then the input is generally a voltage source. But I've seen old designs where the 2nd stage uses a diff pair, where the emitters connect to signal ground w/o large emitter resistors or constant current sinks/sources, and the transfer function is expressed in beta1/2 terms. A diff amp, like a single amp, has a voltage gain as well as current gain.
Class B amp distortion proves what? It proves that if Vbe were its ideal zero value, said distortion would not exist. A b-e junction is a forward biased diode, which in the ideal has a forward current with ZERO forward voltage drop. If Vbe were zero, then crossover distortion is gone. Many who dispute current control model insist that Ib is a defect, yet it is not apparent to them that Vbe is just as much a defect as any other undesirable quantity.
Current mirrors do not work well if design relies on 2 devices matching in Vbe. Even identical geometries on the same substrate do not always balance since a slight inequality in current will heat up the device with larger current more than the other, and that device eventually hogs the current. Fortunately, some intrinsic emitter resistance, r_e, is present and corrects this. What emitter resistance does is force the 2 devices to have equal Ie values. If Vbe's mismatch, but an external Re is present, it will force Ie's to closely match, even if the Vbe's do not. I've designed many many mirrors, and I can achieve great tracking/matching when the 2 or more devices are different part numbers, on different boards, different temp. I use enough emitter degeneration resistance and force all parts to have nearly identical Ie values, even though Vbe values do not match well. As long as the devices have near equal Ie, all the Ic values are alpha times their respective Ie. If all devices vary in temp and differing parts, even if beta varies from 80 to 400, alpha will vary from80/81 to 400/401 resp., which computes to 0.9877 to 0.9975 resp. Despite beta ranging 80 to 400, collector currents vary by 1%.
Finally, translinear loops merely demonstrate that the transconductance of a bjt, "gm", is directly related in a linear fashion to the dc collector current, i.e. gm = Ic/Vt. The fact that a bjt possesses transconductance does not make it voltage controlled. A bjt has both transconductance and current gain, gm and beta. Likewise a FET (J- or MOS-), has both transconductance and current gain. But possessing a current gain does not make the FET current controlled. The bandwidth limit for a FET is described either with "ft" the transition frequency, or input capacitance. The transition freq, ft, is that freq where the current gain has dropped to unity. I.e. at ft, id/ig = 1. Above this ft value, the FET has a current gain less than unity, so that it is less useful as an amplifier. Or, it is common to use input capacitance to compute gate current. At some freq, the gate current will reach the value of the drain current, and current gain is unity, the device has reached its bandwidth limit.
I will elaborate if desired. Please study this carefully before asking questions, thanks.
Claude
