Claude Abraham
Member
Very good job Mr. Al. I applaud you for your patience. You have more than I have, that's for sure.
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Very good job Mr. Al. I applaud you for your patience. You have more than I have, that's for sure.
Claude,
MrAl certainly deserves a thank you, but you neglected to mention Jony130 for his contribution. And do I not also deserve an honorable mention for not walking away, and agreeing with something I did not quite believe just to settle the matter? Unfortunately my unfamiliarity with Ltspice and a bum transistor chart did not help to shorten the discussion either.
MrAl,
I said that I agreed with you about Ic controlling the transistor, especially during saturation. That is certainly true, but I was wondering why you made that point. It was never in doubt, was it? Much of the discussion was whether Ic increased enough to bring the transistor out of saturation. And while in saturation, Ib doesn't have much effect on Vce(sat), but it does on what value Ic brings the transistor out of saturation. So it can be said that Ib also controls the transistor during saturation at the same time. I don't want to start another round of discussion on this, because I believe we each see the other's point of view.
Ratch
One salient point I'd like to make is that earlier on, you mentioned that the I-V curve of a bjt cease to be exponential when the device goes into saturation. Although that is true if we view only Ic vs. Vbe, keep in mind that there are 2 things going on. Ebers & Moll pointed this out in their 1954 paper. In saturation, the b-c jcn is forward biased. The base current has 2 components, base to emitter, & base to collector. In the base region these 2 components add together. But the collector current Ic, is made up of the component due to the b-e jcn forward bias, & the other component, due to the b-c jcn forward bias. The 2nd component subtracts from the 1st. If we visualize an upside down bjt, the collector becomes the emitter & vice-versa.
The relation between Ic & Vbe is still logarithmic, even in saturation, but Ic/Ie consists of 2 parts in opposition.
Ic = alpha_n*Ies*(exp(Vbe/Vt)-1) - Ics*(exp(Vbc/Vt)-1),
Ie = Ies*(exp(Vbe/Vt)-1) - alpha_i*Ics*(exp(Vbc/Vt)-1),
where alpha_n/alpha_i are the normal/inverse common base current gains resp., & Ies/Ics are the reverse saturation scale currents for the b-e/b-c jcns resp.
In the active region, the 2nd term involving Ics/Vbc is small since the b-c jcn is reverse biased, i.e. leakage current. Hence 1 term is all that is needed, & it exhibits exponential I-V. In saturation, 2 terms are involved, & the exponential nature of the junctions is not immediately apparent.
rt of as,
I hope I convinced you that both the transistor and vacuum tube to not use resistance to modulate a signal.
Ratch
Hello,
It sounds almost like you are saying that the transistor does not dissipate energy.
...
That's where the controlled resistance interpretation comes from. Im sure you are aware of this.
I think if new people to electronics stop thinking of transistors as amps and more like switches there would be a lot less confusion.
A transistor controls a higher power, Volts, Amps or both to look like the original signal only bigger.It takes a while to convince them it is controlled