How can a transistor amplify current in a circuit?

Question

At 2:57 of this video, it is said that when we apply a small current to the base of the transistor, a large current is passed through. This is fine, but what bothers me is what he says right after: "If we now manipulate the base current in a certain change, the other current changes proportionally with much higher amplitude"

I just can not understand why it is that variations in the base current should amplify the current through the transistor. If I am understanding correctly, if we apply a current on a transistor, then we simply decrease the size of the depletion region and hence make it more conducting. However when we do this, the transistor didn't really amplify any signal, but rather let more of the current pass through.

So, why is it that we say a transistor is working as an amplifier here? Or am I misunderstanding?

exactly, and since the increase in collector current is larger than the increase in bae current, we call that amplification — Marcus Müller, Mar 01 '21 at 11:54
That is adding current but then why we call it 'amplfying'? Amplifying I understood as you have some quantity and you scale it up by some factors — Reine Abstraktion, Mar 01 '21 at 11:56
It is not adding current, it is a multiplication. Example for a current amplification of 100 (\$\beta\$ = 100): Base current = 1 uA => Collector current is 100 uA; Base current = 2 uA => Collector current is 200 uA. — Bimpelrekkie, Mar 01 '21 at 11:57
Where are the extra micro amperes coming from? @Bimpelrekkie Is it from transistors charge carrier or because the resistance of the transistor device is reduced? — Reine Abstraktion, Mar 01 '21 at 11:59
Amplifying is where we have more power in the output than the input; if there is more current in the collector than the base, then *even if the variation of voltage at base and collector were the same*, we would have amplification. — Peter Smith, Mar 01 '21 at 12:05
*Where are the extra micro amperes coming from?* From another (part of the) circuit or a sensor. To understand amplification, it doesn't really matter. **How** circuits work (at electron level) is difficult so don't worry if it all doesn't make sense yet. If you keep learning then one day it will all come together and you'll understand. Also, you can use transistors fine without understanding **how** amplification works. — Bimpelrekkie, Mar 01 '21 at 12:55
"we simply decrease the size of depletion region and hence make it more conducting" in other words we modify its resistance... the word "Transistor" came from "Transfer" and "Resistor". That change in resistance allows a much larger current variation than the base current variation and that is the amplification. — , Mar 01 '21 at 13:00
.....yes, and the size of the depletion region can be increased/decreased by changing the base-emitter voltage Vbe around the nominal bias point of Vbe=0.65...0.7 volts, As a consequence. also the current Ib changes (but this is a secondary effect) — LvW, Mar 01 '21 at 13:10
What's the difference between "letting more pass through" and "amplifying the signal"? If a small current flow makes a large current flow, that's amplification. — user253751, Mar 02 '21 at 09:31
How is this not a duplicate? After [more than 11 years](https://electronics.stackexchange.com/questions/1/arduino-stepper-motor)? — Peter Mortensen, Mar 02 '21 at 10:01
I'm sorry but how does that linked question relate with mine? @PeterMortensen — Reine Abstraktion, Mar 02 '21 at 10:44
Peter Mortensen, yes - you are right. And that is a very surprising situation. In spite of many counter examples and theoretical as well as experimental observations, there are still many people believing (believing !!) on current-control. Its a kind of religion and very often a technical discussion is useless. — LvW, Mar 02 '21 at 11:31
@PeterMortensen er... you linked a question on Arduino and stepper motors... :-) LwW, perhaps you might want to techically discuss where Streetman is wrong. — Sredni Vashtar, Mar 03 '21 at 14:20
@SredniVashtar The linked question is the oldest question on the site (id 1), I think PeterMortensen was making a point about the entire set of questions on this site, not anything to do with relevance to that question specifically. — user1937198, Mar 03 '21 at 17:24
@user1937198 OOOOHhhhhhhh, I see now :-) (blushes). At any rate, sure this is a duplicate. Here's one: https://electronics.stackexchange.com/questions/101595/basic-operation-of-a-bipolar-junction-transistor?rq=1 . I can recall at least other two instances but I cannot locate them now — Sredni Vashtar, Mar 03 '21 at 17:29

LvW · Accepted Answer · 2021-03-03T18:35:18.157

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I can understand your doubts - because, in reality, the transistor does NOT amplify the base current. It is true that the collector curent \$I_c\$ is proportional to the base current \$I_b\$ (\$I_c/I_b=\beta\$), but this is a kind of correlation.

(It is really a pity that there are still some books and publications claiming that \$I_b\$ would determine \$I_c\$. Nevertheless, during design and/or analysis of transistor stages we can in many cases treat the transistor as if \$I_b\$ would determine \$I_c\$; this is because the relation \$I_c=\beta I_b\$ does apply - but it is a correlation and does not reflect a causality).

It is not a problem to show and to verify that \$I_b\$ as well as \$I_c\$ are both dependent on the voltage \$V_{be}\$ according to Shockley's equation \$I_e = I_s[\mathrm{exp}(V_{be} / V_t)-1]\$ because the emitter current \$I_e\$ is split into a very small current (\$I_b\$) and a larger current \$I_c\$ (\$I_e = I_b + I_c\$).

Final (summarizing) statement (with respect to the long list of comments):

The following sentence alone cannot explain the transistor principle, but it shows that it is the VOLTAGE which plays the decisive role:

In every conductor/semiconductor, a current (movement of electric charges) can exist under the influence of an electric field only. In electronic circuits, this E-field is generated by an external voltage. A change of the voltage (of the E-field) causes a change of the current. This is true, of course, also for diodes, bipolar transistors (BJTs) and FETs.

edited Mar 03 '21 at 18:35

answered Mar 01 '21 at 12:53

LvW

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Thank you kind sir, I was almost losing it after spending one whole day reading from different books and seeing videos and not understanding. Thank you again. – Reine Abstraktion Mar 01 '21 at 13:22
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Buraian, of course, there are many good (and serious) books/publications which gives you the correct information how transistors really work ("Art of Electronics", papers from Berkeley, Harvard, Stanford, MIT,...). But it is realy disappointing that there are so many contributions (printed and internet) which simply state "current controlled" without any verification. – LvW Mar 01 '21 at 14:30
Buraian, here is a source I can recommend: https://ncert.nic.in/ncerts/l/leph206.pdf – LvW Mar 01 '21 at 16:30
Ahh thank you but that brings me full circle :P, that book was is the textbook prescribed for highschoolers who had taken physics in India, I found it a bit difficult to understand hence was trying to find simpler explanations online – Reine Abstraktion Mar 01 '21 at 16:56
But - for my opinion - it contains a good and detailed explanation how the BJT works in principle. Be careful with online presentations....they can seldom replace a good textbook. – LvW Mar 01 '21 at 17:07
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So, you have come to the conclusion that Streetman - who wrote a full solid state devices book -is wrong when he shows how the base current controls the collector current? – Sredni Vashtar Mar 01 '21 at 20:44
@SredniVashtar William Shockley himself didn't understand BJTs ( that's why Ebers-Moll could become famous. ) Perhaps Shockley's was a strategy to avoid having their patents again rejected by USPTO, since (like vacuum tubes and FETs,) the BJT is a voltage-input device, and the FET had been patented two decades earlier. Therefore, the BJT absolutely HAD to be "current-input," even though it actually wasn't. Then, generations of textbook-writers make the same mistake: ignoring Ebers-Moll description, and instead spreading Shockley's mistake. – wbeaty Mar 02 '21 at 00:30
@wbeaty in the early days of the transistor, it was perfectly understandable not to understand how they worked. Brattain, Bardeen and Shockley broke new ground and the physics was not not well understood. But today it is (mostly :-) ) and there are no more problems with the patents. So, you think that Streetman, a respected author of a solid state devices book is... repeating "Shockley's mistake" in those handful of pages of his textbook? Where exactly is he committing the mistake? – Sredni Vashtar Mar 02 '21 at 00:37
@SredniVashtar Also note: Ib determines Vbe (see diode equation and its explanation.) Then next, the EB junction-barrier determines Ic. In other words, we can hide the Vbe diode-physics, and teach instead a simplified version where Ib controls Ic. In truth, Ib -> Vbe -> Ic where Vbe determines the potential-barrier, which directly controls Ic. Then Ib is only a non-ideal and undesired leakage current, analogous to gate current of FETs. (When hfe > 50, we easily design BJT circuits where the presence of Ib is tolerated, while its small value is not relevant to circuit function.) – wbeaty Mar 02 '21 at 00:43
@wbeaty you did not read Streetman, did you? ;-) – Sredni Vashtar Mar 02 '21 at 00:48
@SredniVashtar I don't have Streetman copy (during BSEE, my text was IIRC ?Chang? device-physics, not circuit equations.) Again: does Streetman use Ebers-Moll for transistors, explaining hfe by applying diode concepts (potential barrier height etc.?) If Ebers-Moll is missing, then Streetman isn't teaching BSEE courses. Engineers must learn the internal BJT physics: Ebers-Moll, Gummel-Poon, and not Shockley's mistake of "current controlled" BJT, wrongly focused on "alpha" and "beta" in 2-port networks. Open the "black box" and we find Ebers-Moll, where no mechanism lets Ib directly alter Ic – wbeaty Mar 02 '21 at 01:27
Sredni Vashtar, in one of your former comments I read "Steetman...shows how the base current controls...". Did he really "show" and verify HOW such a control will happen - or was it just a statement without any evidence? Anyway: Is Streetmans book the only ref. you have? What about papers and university lessons from Berkeley, Harvard, Stanford, MIT, Arizona Univ.....? – LvW Mar 02 '21 at 08:53
And - not to forget: I alsao rely on my own observations/interpretations: Re-feedback, EARLY voltage, voltage gain (transconductance), Vbe-tempco, BJT operational conditions (A/B/C),.. – LvW Mar 02 '21 at 11:28
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We can take Hill [at his own words](https://cr4.globalspec.com/comment/720033/Re-Voltage-vs-Current) – J... Mar 02 '21 at 15:31
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"in reality, the transistor does NOT amplify the base current" Consider a microphone that produces a small peizoelectric current, which is then fed into the base of a transistor, which has a large constant power source applied to its collector. The current that goes out of the emitter will be proportional to the peizoelectric current generated by the microphone, but significantly stronger. It's turning the constant input into the collector into a varying output out of the emitter. – nick012000 Mar 03 '21 at 04:54
nick012000, nobody will deny the proportionality between base current and emitter current. – LvW Mar 03 '21 at 09:18
"We can take Hill at his own words –"....yes, this is in accordance with his book (co-author): Art of Electronics. – LvW Mar 03 '21 at 09:35
What's the point in adding another reference, if you refuse to read them? I gave you Streetman, Levinstein & Shimin, Senturia, Eisberg & Resnick, even Hu. Want another one? Pierret "Semiconductor Device Fundamentals": "Control of the larger IC by the smaller lB is made possible by the coupled junction arrangement that physically divides the small electron and large hole currents crossing the EB junction into two separate current loops". Another one? Mishra & Singh "A change in the base current alters the minority carrier density in the base and causes a large change in the collector current" – Sredni Vashtar Mar 03 '21 at 13:47
You keep reverting back to the straw man argument but saying that you can't disentangle V from I does not mean that the BJT is exclusively current controlled. It can be seen as either, no matter what Hill says. He is adopting the point of view of a circuit designer: the Ebers-Moll equation Ic = f (Vbe) implicitly carries more information of Ic = beta Ib because the exponential dependencies in the latter one are hidden. And i gave you at least four or five solid state device and solid state physics books that confirm how the carriers injected in base influence the value of Ic. – Sredni Vashtar Mar 03 '21 at 14:14
Sredni Vashtar, I also could say "What's the point in adding another reference" (in fact, more references). I also could add (and I have done this already) a lot of references pro pure voltage control. But this does not help at all. More important: What are your comments to the PRACTICAL examples I have mentioned several times? Therefore: Please, give me one single example circuit which can be explained in its function with current-control only! In contrast, I have mentioned several effects/circuits which can be explained only with voltage control. No comment up to now... – LvW Mar 03 '21 at 14:31
@LvW Indeed - this was in support of your assertion. AoE does present the current-input model of the BJT, of course, because it's a complete academic text, but I think this statement from Hill show his view that the current-input model is nothing more than just a somewhat useful model and isn't really to be taken as a physical interpretation of the device's operation. – J... Mar 03 '21 at 15:25
@J..., at the end of chapt. 2.09 in AoE we can read: "Clearly, our transistor model is incomplete and needs to be modified .....Our fixed-up model, which we will call transconductance model, will be accurate enough for the remainder of this book". (End of quote). And at the beginning of chapt.2.10 (quote): "But to understand diff. amplifiers....and other important applications you must think of the transistor as a transconductance device - coll. current is determined by base-to-emitter voltage". – LvW Mar 03 '21 at 15:39
@LvW Yep, agreed completely. I added the link as a backstop to the argument that because AoE contains and presents the current-input model that it has some validity as a physical model. – J... Mar 03 '21 at 15:41
@LvW why on earth you keep asking to provide an example of 'current control only' when all I have been repeating is that you cannot disentangle V from I and hence you can have both (maybe more cumbersome to describe). And besides, do you really think that the (appr.) independence of gain from beta due to Re degeneration cannot be explained when using ib instead of vbe? Temperature dependence is right there when you consider logs of I instead of exps of V. PTAT thermometers are based on that. – Sredni Vashtar Mar 03 '21 at 15:51
Of course, voltage and current are closely related, but that does not mean that we are not able to decide whether a certain effect is caused by voltage (e.g. FET) or current (e.g. magnetic field). No current without driving voltage. Both are not interchangeable! – LvW Mar 03 '21 at 16:11
Everywhere in natural sciences the principle of causality prevails - with few exceptions, all effects to be observed can be clearly explained and assigned to a certain cause (physics, chemistry, meteorology, mechanics,...) . Do you really believe that the BJT would be a great exception to this and is such a mysterious element that an unambiguous explanation of its working principle (voltage- or current-controlled) would be not possible? All the observed effects and also the common practice of circuit design speak against it. – LvW Mar 03 '21 at 16:12
Yes, voltage and current are such an exception - they are concomitant. In a conductor - ie in the circuit around the transistor - you can't have one without the other. This is basically due to the fact that the moving charges that make up the current carry with them and electric field that would alter the voltage if there was not a resinstatement of the charge displacement. Inside a semiconductor you CAN HAVE current without voltage!!! Ever heard about diffusion? The Haynes-Shockley experiment? But you did not answer my question about your reiteration of Re degeneration as proof of V-control. – Sredni Vashtar Mar 03 '21 at 17:08
I did not answer this question because I NEVER have mentioned Re degeneration in the context you have mentioned ("independence of gain from beta due to Re degeneration".) So your quotation was not correct. My argument was (and still is): The voltage gain with a simple common-emitter or common-base stage (without feedback) is determined by the transconductance only (gm*Rc). The transconductance is gm=d(Ic)/d(Vbe). It tells us how much the current Ic changes for a change in Vbe.Under the same DC conditions the gain does not depend on beta. I am sure that this is not a new information to you. – LvW Mar 03 '21 at 18:05
However, as you have mentioned the case of Re-degeneration: It is quite clear that such a resistor Re provides current-controlled VOLTAGE feedback. Only in such a case the input resistance goes high (system therory). Hence, this is a clear evidence for voltage control. – LvW Mar 03 '21 at 18:47
Sorry if possibly stupid, but suppose I drove a time varying DC voltage having some wave shape, now when this pass through the transistor would this get wave form get scaled up by some factor or you add some constant? If my understanding is right, then I think the adding constant is correct. @LvW – Reine Abstraktion Mar 03 '21 at 22:24
Buraian, A "time-varying DC voltage" will not "pass through the transistor". Such a voltage variation will cause the same effect as a sinusoidal signal - why should there be a difference? Such a time-varying signal "delta-Vbe" will cause a corresponding "delta-Ic" (and a corresponding "delta-Vout"). The same effect can be observed when selecting a suitable DC-bias point: Vbe=0.7 volts will cause a larger quiescent current Ic than Vbe=0.65 volts. In you words: The waveform of a changing dc signal will be "scaled up by some factor" (the voltage gain). – LvW Mar 04 '21 at 08:12
Buraian, only now I can imagine what you mean with "...add some constant". Yes - the mentioned delta values (see my former comment) ride upon the DC quiescent current Ic or DC output voltage, respectively. In contrast to sinusoidal signals we cannot separate both parts (DC and arbritrary waveform) using a capacitor at the output. – LvW Mar 04 '21 at 13:24

score 9 · Answer 2 · answered Mar 01 '21 at 12:51

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I presume you are just starting out in electronics and, as such, the WHY of the mechanism of a transistor's behavior can be quite intimidating. It's rooted in the physics of how charges work inside the structure that has been created inside the device. If at some point you choose to pursue electronics at a university, perhaps an Electrical Engineering degree program, you will eventually (likely as a 3rd or 4th year student) reach a class named "Semiconductor Physics" or perhaps "Solid-state Physics". There you will learn, in depth, what's going on.

But until that point I urge you to do what engineering student and the vast majority of transistor users do, take it on faith that it works and focus on its higher-level behavior in a circuit rather than the physics of its operation.

If you really want to know this today, I applaud you for your interest and there are any number of resources that you might study to gain an understanding of how semiconductors and ultimately a bipolar transistor works.

Here is a link to an introduction to transistors article that I found that may help you get started:

Transistors

Here is a short excerpt from that article:

answered Mar 01 '21 at 12:51

jwh20

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Thank you , that article and linked seems helpful for my studies. – Reine Abstraktion Mar 01 '21 at 13:25
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Quote: "But until that point I urge you to do what engineering student and the vast majority of transistor users do, take it on faith that it works...". But - dont you think that one should know HOW it works? How can somebody design a circuit without knowing how the most important part (BJT) works in principle? I think, its not too complicated....when you know how a pn junction behaves (a simple pn diode) it will not be a great problem to accept that the majority of the emitted electrons will travel to the collector - thats all! – LvW Mar 01 '21 at 14:39
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Clearly you didn't read the rest of my answer. But again, how many people regularly and successfully use transistors without any clue of the physics behind them? For the vast majority of users, the small signal model is more than enough. – jwh20 Mar 01 '21 at 14:50
Well I agree that for "succesfully use transistors" it is not necessary to have a deep knowledge on transistor physics. However, my main point is (and THIS is the core of the original question): How can we control the amount of collector current? And - at least - every user should be aware that it is solely the voltage Vbe that determines the collector current. Otherwise, he will continuously observe contradictions (emitter follower, current mirror, diff. amplifier, Re-feedback principle, EARLY effect, class-B operation,....) – LvW Mar 01 '21 at 15:39
@LvW But... the Early effect is an example of a place where collector current isnt just a function of Vbe. It is NOT "solely the voltage Vbe that determines the collector current". – Matt Mar 01 '21 at 20:07
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@Buraian Beginners do grasp the simple basics of FETS. The BJTs cause all the trouble! We haven't answered their questions if we say "just use it, without understanding," or say "its just a bunch of equations." (The same applies to FETs!!!) Why explain FETS, but fail explaining BJTs? It's because the simplified BJT explanations for beginners, are simply wrong: they are widespread misconceptions which have invaded some textbooks. Base current cannot influence Collector current. Instead the accurate explanation involves basic diode operation, and changes to the potential-barrier (Vbe.) – wbeaty Mar 01 '21 at 23:57
@wbeaty So, you too believe that Streetman is wrong? In my answer here - in the reference section - there is an excerpt about the point he makes about the role of base current in controlling collector current: https://electronics.stackexchange.com/questions/547625/working-of-bipolar-junction-transistor-with-electron-flow/548094#548094 Namely: "iB can indeed be used to determine the magnitude of iC." and "Clearly, the supply of electrons through iB can be used to raise or lower the hole flow from emitter to collector." – Sredni Vashtar Mar 02 '21 at 00:05
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@SredniVashtar Instead trust AOA Art Of Electronics, which gets the BJT physics right. If Streetman doesn't employ Ebers-Moll and EB potential barrier, then Streetman is teaching oversimplifications not appropriate for engineering classes. Think: inside FETs, the Ig gate-leakage current can indeed be used to control the magnitude of Id (a Vgs/Ig proportionality does exist.) Even in vac tubes, the grid current can indeed be used to control the plate current. But if we offer these as an *explanation* of FET or vac-tube operation, we're permanently damaging our students' understanding. – wbeaty Mar 02 '21 at 01:04
@wbeaty It seems to me you are bringing a knife to a gunfight. I mention a solid state devices book, written by a respected author in his field who chose to devote several pages of a devices textbook to explain why and how base current can control collector current, and you bring TAoE, a circuital solutions cookbook (of the kind focused on the two-ports representation mentioned above) that devotes, what? Half a column? to handwavingly describing the physics of the transistor. I like TAoE, but I would not take it as the gospel in solid state physics. -continues- – Sredni Vashtar Mar 03 '21 at 13:41
--- In my eye what Horowitz and Hill are saying is that the Ebers Moll equations ic = f (vbe) sums up information that is not explicit in the ic = f (ib) equation (because in the latter the exponential dependence has 'canceled out' so to speak), and that is why it's better to use that representation. But this does not implies there is a one-way causation link - for example when designing PTAT thermometers you can also use the inverse relations for the junction - where voltage depends logarithmically on the current. – Sredni Vashtar Mar 03 '21 at 13:41
@SredniVashtar first understand how BJT transistors actually function, and then Ebers-Moll and Gummel-Poon models will seem simple and obvious (and major textbook flaws become visible!) Base current cannot directly influence collector current. Again, it's not complicated: by simple diode physics Ib can determine Vbe. Then Vbe and the e-fields within EB junction determine Ic. That's the essence of "transisting." Some popular textbooks are flawed, and they pretend that Ib somehow influences Ic, while never mentioning that BJT are voltage-controlled devices, similar to FETs and vacuum tubes. – wbeaty Mar 04 '21 at 01:35
@SredniVashtar the AOA text (and especially the AOA lab manual) in detail shows why we cannot use hfe, since our designs will fail. I had to learn this the hard way, wasting much time during my first professional analog designs. "hfe-think" is like brain-damage! That's the reason why AOA teaches us to stick with Ebers-Moll, and with BJT designs based on voltage-input, where hfe is irrelevant. That Ebers-Moll also contains the correct explanation for bjt physics, while hfe does not, is a separate issue. Author of AOA says: https://cr4.globalspec.com/comment/720374/Re-Voltage-vs-Current – wbeaty Mar 04 '21 at 01:48
@wbeaty You are confusing the rules of thumb of a circuit cookbook with the physics of a devices textbook. We're not talking about making a design independent of beta or T, but of the (non-existent IMO) difference between using Ib = EXP (Vbe) vs Vbe = LOG (Ib). We are talking about Ib and Vbe (or Ie and Vbe) being concomitant and not the cause one of the other. You CANNOT alter - and mantain - Vbe without concomitantly supplying charge via Ib. That is the point. Ib is not a defect - it is a necessity no matter how much you, LvW and Gilbert consider it a nuisance. Even Hu makes that point. – Sredni Vashtar Mar 04 '21 at 23:07
And Streetman shows why, based on charge neutrality argument. (And of course Streetman treats Ebers Moll and Gummel Poon). I have yet to hear any objections against the charge neutrality argument Streetman explicitly makes in his solid state device book. He's not handwaving a description like Senturia - he details why the tiny little Ib can make a larger Ic flow. According to you he must be obviously wrong, so where is he wrong? LvW is at least consistent in his objection: he firmly believes that voltage causes current, period. But I am a bit confused about your stance [...] – Sredni Vashtar Mar 04 '21 at 23:09
[...] you say that current causes voltage ("Ib determines Vbe") and "THEN next" voltage causes current ("Vbe and the fields determine Ic"). Does "THEN" mean 'at a later time?' You want to see cause-effect? Make time explicit, then we'll talk. Instead of voltage (or current) revert back to charge and then we will see what causes what. And if you are advocating charge control, then Hu shows how Ib can determine Ic. – Sredni Vashtar Mar 04 '21 at 23:09
Quote: "LvW is at least consistent in his objection: he firmly believes that voltage causes current.." (End of quote). No - I do not "believe". It is a physical fact that movement of charged carriers forms the quantity we call "current". Why do they move? Because they "feel a corresponding force". This force is caused by an E-field within the conducting materal. And this E-field is caused by the applied voltage. So - can we have a current without voltage within the transistor? No. We cannot "inject" a current - thats "labor jargon". Please, in your response, tell me if something is wrong. – LvW Mar 06 '21 at 09:44
Quote (Sredni Vashtar):"Hu shows how Ib can determine Ic." (End of quote). Sredni Vashtar, I am afraid, you are mixing here models with physical description. Here is what Hu (Berkeley) says in chap. 8.3: "Ib is an undesirable but inevitable side effect of producing IC by forward biasing the BE junction". – LvW Mar 06 '21 at 09:53

Circuit fantasist · Answer 3 · 2021-03-05T21:28:23.673

In fact, OP has understood the naked truth about the transistor "amplifier"... that it is not an amplifier at all... on the contrary, it is an "attenuator". At this stage, OP does not need detailed explanations; he needs confirmation of his guesses.

It is considered the transistor is an active element used to build amplifiers... but IMHO this is not true. The transistor is not active but passive element; the only thing that it can do is to dissipate power. So, the transistor is not amplifying but attenuating element. It is just a "resistor" (non-linear, electrically controlled but still a resistor) that decreases the current.

The true amplification is impossible; so there are no real amplifiers. The so-called "amplification" is just an illusion, a clever trick... and the "amplifier" is just a "magic box" where we see higher output power... but this is not the amplified small input power. This is else's power... of the supply source.

In analog electronics, we implement such an "amplification" in the possibly most paradoxical, absurd and silly way - to obtain output power higher than the input one, we get a big power source and then throw away a part of it (from zero up to the whole power). In comparison, in energetics, they can not afford to do it.

I use this approach in classes with my students to clarify such vague definitions of amplifier as "electronic circuit that uses electric power from a power supply to increase the amplitude of a signal applied to its input terminals" (Wikipedia). For example, here is a seminar in 2004, in which we discussed the philosophy of the transistor "amplifier". Another Wikibooks story from 2008 describes how my students studied this phenomenon in the lab in order to reinvent the BJT current mirror.

In 2013, I asked a similar ResearchGate question which provoked a heated discussion. I hope it will be useful to you.

Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/120382/discussion-on-answer-by-circuit-fantasist-how-can-a-transistor-amplify-current-i). — Voltage Spike, Mar 03 '21 at 18:29
BTW, the term "amplifier", used without qualification, implies power amplification. This does not mean that the terms "current amplifier" and "voltage amplifier" are invalid or not used. www.britannica.com "Amplifier, in electronics, device that responds to a small input signal (voltage, current, or power) and delivers a larger output signal that contains the essential waveform features of the input signal." — Elliot Alderson, Mar 05 '21 at 22:14
@Elliot Alderson, I have nothing against using the terms "current amplifier" and "voltage amplifier"... and I regularly use them... but I want us not to forget what lies behind them. The purpose of my answer is to reveal the idea of what we call "amplification" by showing that, at first glance, it is absurd but inevitable in analog electronics. The absurd thing is that **we get gain through attenuation**: we attenuate the voltage of the supply voltage source but still this voltage is many times higher than the input voltage... and we say that this is an amplified input voltage. — Circuit fantasist, Mar 07 '21 at 02:29

score 7 · Answer 4 · answered Mar 01 '21 at 14:24

At 2:57 of this video, it is said that when we apply small current to base of transistor, a large current is passed through,

The ratio, large collector current to small base current, stays more or less constant. It's sufficiently constant that we give it a name, β, or Hfe, or 'current gain'. This is the main defining feature of a transistor, it's what a transistor does. If the base current increases or decreases by a_bit, then the collector current increases or decreases by roughly β*a_bit. That, and the broadly equivalent base voltage controlled collector current behaviour, is all that 99% of practicing electronics engineers use in their day jobs.

Don't worry too much about what depletion region is doing until you're ready to take a full course on Semiconductor Physics. It doesn't really help to know a little bit of the story, enough to get you worried and drawing the wrong conclusions

score 6 · Answer 5 · edited Mar 01 '21 at 14:14

Keeping it in simple terms for the beginners that we all were once...

The transistor is not an amplifier. On its own, it does absolutely nothing.

However, the transistor can be used to build a circuit that is an amplifier.

For a BJT amplifier, that circuit will use the transistor to pass a current from the circuit's power supply to to the transistor's output pin. That output pin current will be proportionally larger than the current through the transistor's input pin.

This is because the transistor not a 'passive component', it's an 'active component' i.e. it requires a power supply to perform its function.

score 5 · Answer 6 · answered Mar 02 '21 at 09:37

, if we apply a current on transistor, then we simply decrease the size of depletion region and hence make it more conducting, however when we do this , the transistor didn't really amplify any signal but rather let more of the current pass through.

So in other words, a small current flow causes a large current flow.

That's amplification.

It doesn't say that the large current has to come from the same place as the small current. No, the large current just comes from some kind of power supply. That fact doesn't mean it's not amplification.

KD9PDP · Answer 7 · 2021-03-01T13:36:12.490

To understand what's happening, you have to apply a little bit of quantum mechanics and classical statistical mechanics. When you apply a voltage to the base of an npn, you do shrink the base emitter depletion region, thereby exponentially allowing more electrons in the emitter to diffuse into the base, which then transit the base and end up coming out of the collector. You also decreased the size of the base-collector depletion region, allowing exponentially more holes to flow from the base to emitter.

The net sum of all of these effects is the appearance that a small base current is proportional to the larger collector current (in active mode). The amount of gain depends on how fast and efficiently the minority carriers (electrons in the base of an npn) diffuse through the base from the emitter to the collector. That's the "base transit time" term you see in bjt equations. The faster that is, the higher beta is.

So it's a little more complicated that just two diodes back to back.

How can a transistor amplify current in a circuit?

7 Answers7