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Assistance with Transistor Theory

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Laplace

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...whether IB causes IC...
Now go back to #110 with the current driven base. If you vary the base current, then what happens to the collector current? There is no 'external circuit' there because the resistance of a current source is infinite, i.e., the BJT sees an open circuit at the base along with a forced value of base current. As the collector voltage is varied from zero to maximum, what happens to the collector current? What happens at the base?
 

KrisBlueNZ

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Sometimes I think that there is a Vbe of about 0.7V, or 0.6V, or about half a Volt, but a lot of the time, if the supply is 9V or 12V or 15V or more, I totally ignore Vbe. And I'd always assumed Vbe was FIXED at whatever the correct value is - though I now know from your comments is wrong. And this seems to work for the simple circuits I build. (*)
So the question I would like to ask of all readers is, "have you EVER thought about setting Vbe precisely or thought about delta Vbe in designing a circuit?"
Yes, I consider VBE often. In some circuits you can't ignore it - an emitter follower, or a complementary output stage in an audio amplifier which requires a base-to-base voltage drop to set the quiescent current.

I consider it even in simple cases such as when a circuit is driving a transistor with its emitter at 0V using a base voltage divider. I choose the ratio between the resistors so that the base voltage reaches 0.7V when the voltage from the driving circuit is around 1/3 to 1/2 of the supply voltage (or some other voltage if the driving circuit doesn't swing widely).

I don't consider VBE variations - just because it doesn't vary that much - except again in output stages where variation due to temperature must be compensated to avoid thermal runaway.
I guess there are some pro designers here who would say yes, but I can hand on heart say, "No, never" and I'd bet that's true for most amateurs and therefore for people like OP who just want to know how a transistor WORKS (in a circuit.)
I don't know about that. I think most people learn pretty early on that you need about 0.6V between the base and the emitter before anything starts to happen.
 
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KrisBlueNZ

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Well the fact that without a voltage you wouldn't have a current would you?
True, but that's true of any non-reactive component. A resistor, a light bulb, a varistor, an LED. So pointing it out doesn't tell us anything.

Without voltage, you couldn't make an LED light up. But LEDs are described as current-controlled because their brightness is roughly proportional to the current, and not really proportional to the voltage.

I agree with Ratch that whether it makes sense to describe a transistor as current-controlled or not depends on the internal mechanism of its operation.
 

KrisBlueNZ

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Now go back to #110 with the current driven base. If you vary the base current, then what happens to the collector current?
Obviously, it varies. The collector current is roughly proportional to the base current. I understand that! What interests me is Ratch's explanation about how base current occurs. As I understand it (which may not be very well), he describes it as an unnecessary by-product of the way a transistor operates internally, and is best described as being a roughly constant fraction of IC.

You didn't answer my direct question that I asked in post #92 and clarified in post #116, so I'll ask another - for anyone to answer please.

Why is it impossible to make a BJT whose base current is less than about 0.1% of its collector current? (In other words, with a beta of more than about 1000.)
 

Laplace

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You didn't answer my direct question...
I thought that I had answered your question. Now I'm disappointed that the answer was not satisfactory. This is the first time I've seen the theory that collector current causes base current, although the Ebers-Moll model does include the concept of reverse current gain but that must be something different. I had hoped someone would perform the tests described for operating the BJT with a current source at the base. The observed behavior might provide the resolution to this issue.

Regarding the next question, I searched for 'super beta' and was overwhelmed with offers to solve my prostate problems. Eventually though I found the following two interesting items:

Super-Beta or high current gain NPN transistors -- In certain analog ICs it is necessary to have very high input impedance and very low base currents. For such applications, the typical current gains of an integrated NPN transistor are not high enough. It is possible to increase the current gain of an NPN transistor significantly by improving the base transport efficiency. In this case the base is very narrow (a few hundred angstroms or less). The collector to emitter breakdown of a structure like this is relatively low (2V-3V) because the collector base depletion layer can punch through the active base region into the emitter. This is the punch-through or "super-beta" transistor. Current gains of 5000 are obtainable using this technique at currents of 20uA or so with a Vce of around 0.5V. The fabrication of super-beta transistors in a standard process can be done by using one extra masking step and diffusion. After the base diffusion for the normal NPN transistors a special mask is used to open up the emitter diffusion for the super-beta transistors. At this stage the emitter of the super-beta transistor is only partially diffused. This step is then followed by the masking and n+ diffusion of the standard NPN. Owing to the extra diffusion step for the super-beta transistor, the emitter of the super-beta transistor is diffused slightly deeper than the normal NPN resulting in a narrow base width.
http://signalpro-ain.blogspot.com/2012/01/super-beta-or-high-current-gain-npn.html

Super-Beta (current gain=12,800) results of two-dimensional numerical simulation studies conducted at the Department of Electrical Engineering, Indian Institute of Technology, New Delhi.
INTRODUCTION: Bipolar transistors (BJTs) play an important role in analog circuit design. The common–emitter current gain is one of the important parameters in the design of bipolar transistors. A large current gain can be traded off to improve the speed of the device by decreasing the base width and increasing the base doping. The use of SiGe base is well known for increasing the current gain of bipolar transistors. Recently, a new surface accumulation layer transistor (SALTran) concept has been reported to increase the current gain of the bipolar transistors using a lightly doped emitter and an emitter contact metal with a work function lower than that of the silicon. In this paper, using two-dimensional simulation, we demonstrate that by combining the SiGe base (to enhance the collector current) and the SALTran concept (to suppress the base current), we can realize bipolar transistors with a super beta not reported so far in literature. Our results show that the current gain of the conventional silicon bipolar transistor with a current gain of approximately 40 can be increased to approximately 12800 using the proposed SiGe SALTran structure.
http://www.academia.edu/3172814/A_s...transistor_SALTran_concept_a_simulation_study
 

Ratch

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KrisBlueNZ said:
Why is it impossible to make a BJT whose base current is less than about 0.1% of its collector current? (In other words, with a beta of more than about 1000.)

You will have to ask a semiconductor fabrication man that question. I remember reading about some lab prototyping transistors with high betas, but they could not transfer the method to a production line. Also, secondary effects caused by the modifications needed made the transistors unsuitable for some applications. It is just so difficult to attract charge carriers into the base region by applying a base voltage, and then try to attract all of them into the collector by a collector voltage. Some charge always gets past and into the base.

As an interesting side note. The base circuit makes the charge carriers available, but the collector voltage is needed to remove them from the base slab. But, no matter how high the collector voltage is, if the charge carriers are not present in the base, the current will not increase. You can see this by the flat tops of the Ic-Vce curves found in the data sheets. No current change with collector voltage variation gives the transistor its current source behavior.

Ratch
 
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Ratch

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Laplace said:
I thought that I had answered your question. Now I'm disappointed that the answer was not satisfactory. This is the first time I've seen the theory that collector current causes base current, although the Ebers-Moll model does include the concept of reverse current gain but that must be something different. I had hoped someone would perform the tests described for operating the BJT with a current source at the base. The observed behavior might provide the resolution to this issue.

I believe we said that the base current is proportional to the collector current. Vbe is what causes the waste base current and controls collector current. You mention a model again. Remember, a model simulates what a device does, but not how the device does it. Driving a BJT with a current source is applying an external resistor to the base from the current source, so one would be studying a circuit, not the device alone.

Ratch
 
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Laplace

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Driving a BJT with a current source is applying an external resistor to the base from the current source, so one would be studying a circuit, not the device alone.
But that is just so wrong. A current source has infinite resistance, so driving a BJT with a current source is applying an open circuit to the base and forcing the specified current to flow in the base. With a bias resistor there is a circuit, but with a current source the device is fully isolated and therefore one would be observing the behavior of the device alone.
 

LvW

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But that is just so wrong. A current source has infinite resistance, so driving a BJT with a current source is applying an open circuit to the base and forcing the specified current to flow in the base. With a bias resistor there is a circuit, but with a current source the device is fully isolated and therefore one would be observing the behavior of the device alone.

Hi laplace - I think it is important to note that there is no current source in practice. We always are using a voltage source with a large series resistance Rs.
And what happens if we connect these two elements to the base? We have realized a simple voltage divider with one fixed resistor Rs and one non-linear resistor Rbe=f(Vbe).
And the current adjusts itself to a value which fulfills the condition Ib=Vbe/Rbe (here Rbe is the STATIC resistance of the BE-path other than rbe=hie). And finally, Ic=f(Vbe).
 
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LvW

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I totally ignore Vbe. And I'd always assumed Vbe was FIXED at whatever the correct value is - though I now know from your comments is wrong. And this seems to work for the simple circuits I build. (*)
So the question I would like to ask of all readers is, "have you EVER thought about setting Vbe precisely or thought about delta Vbe in designing a circuit?"

Merlin - I think you are touching the question of dc feedback. Why? Because you mention a FIXED Vbe value.
It would be a very poor design to use fixed Vbe value because
(a) we do not know the exact value for a certain Ic (do you know the exact value for Ib? NO!), and
(b) because of the temperature sensitivity of the BJT.

Therefore, we apply negative dc feedback (e.g. emitter resistor), which DRASTICALLY reduces the sensitivity of Ic against uncertainties (as well as temperature caused variations of) the voltage Vbe. By the way - these effects are responsible for the classical design of a base voltage diovider which is as low-resistive as allowed (due to other constraints like power consumption and signal input resistance).
Remark: It is easy to show that - for sufficient feedback - it does not matter if you are calculating the voltage divider circuit for VBE=0.65 volts or for Vbe=0.7 volts. The resulting Ic uncertainty is in the same order as resistor tolerances.

(*) Can't resist commenting on LzW's comment that he's talking about transistors on their own, not transistors in circuits. I'm pretty sure that transistors don't do anything at all unless they are in a circuit.
The answer is simple: If we want to understand the working principle of a circuit containing a BJT we, of course, must first understand the working principle of the active device.
And all data/graphs given in the data sheets apply to the BJT alone (characteristic curves).
 
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Laplace

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I think it is important to note that there is no current source in practice.
In practice a resistor works well enough and is much cheaper than a good current source. But for making device measurements with constant Ib it is best to use a current source so that Vbe will become its natural value. How else would you get true & accurate device characteristics?
 

LvW

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In practice a resistor works well enough and is much cheaper than a good current source. But for making device measurements with constant Ib it is best to use a current source so that Vbe will become its natural value. How else would you get true & accurate device characteristics?
It was my only point to say that - in fact - we are buiding a voltage divider - as you said: Producing the "natural value" for Vbe.
We shouldn`t deviate too much from our main theme (physical mechanism which detrmines Ic), do you agree?

PS: I`ve just realized, that - up to now - I didn`t get any answer tro my questions contained in my replies'17 and#33. What is your position now?
 
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Ratch

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But that is just so wrong. A current source has infinite resistance, so driving a BJT with a current source is applying an open circuit to the base and forcing the specified current to flow in the base. With a bias resistor there is a circuit, but with a current source the device is fully isolated and therefore one would be observing the behavior of the device alone.

A real world current source does not have an infinite resistance or infinite voltage to drive it. Therefore a current source attached to the BJT base is composed of a very large external resistance and a very large voltage. That setup is not the same as an open circuit where no current at all could be sent into the base. For that reason, I assert that a current source contains an external component (resistor) attached to the BJT.

Ratch
 

Laplace

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That setup is not the same as an open circuit where no current at all could be sent into the base.
The object of determining the device characteristic is to control the amount of base current. If the base resistor is large enough, the base will behave as though the base circuit is open. How large do you think the base resistor needs to be before the BJT can't tell the difference between the resistor and an open circuit? A good current source will have a resistance that large.
 

hevans1944

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Years ago, when I first started puttering around with BJTs (we just called them transistors back when) after years messin’ around with vacuum tube circuits, I recall having a hard time understanding how they worked. I knew the base layer must be pretty thin, the emitter heavily doped to provide charge carriers (electrons for NPN transistors, holes for PNP); the base-collector junction needed to be reverse biased, and the base-emitter junction needed to be forward biased. I didn’t much understand anything else since I had neither training nor experience yet.

So I dreamed up an analogy. The base was like a venetian blind holding back the majority carriers in the base-collector junction. You opened the blinds by forward-biasing the base-emitter junction. The majority carriers in the emitter kinda drifted toward the partly opened blinds and then, once on the other side of the base-collector junction, were accelerated in great numbers toward the collector by the base-collector voltage. In my analogy, some carriers from the emitter were intercepted by the partially opened venetian blind slats and became the base current. I had no explanation for how the venetian blinds were “opened” by the forward bias voltage across the base-emitter junction.

For many years I assumed the base current had something to do with this, that is, the base current was the cause rather than the effect. It’s a good thing integrated circuit op-amps came along a few years later, sparing me (at least temporarily) the heavy lifting math and quantum-mechanical theory that better explained what was really going on.

I still puttered around from time-to-time with discrete transistors, using them mostly as switches since the digital bug had bitten me by then, but with op-amps I could go back to regular analog circuit design using passive components. Well, except for a short digression into using transistor-pairs as temperature-compensated logarithmic circuit elements. Back in the day, you could purchase a matched transistor-pair as an integrated circuit, manufactured on a heated, temperature-controlled substrate. This made them ideal for building log amps as well as front-ends for low-drift differential input instrumentation amplifiers.

There were companies, like Philbrick-Nexus and Burr-Brown, that sold hybrid analog function blocks, a mixture of discrete transistor chips hand-soldered to integrated circuits and discrete components to achieve an analog circuit function, all carefully hidden away inside a block of black epoxy resin with gold-plated leads sticking out the bottom. Very expensive, but they featured guaranteed performance specs and saved a lot of time compared to trying to build it yourself.

Alas (or maybe jump for joy), the analog IC manufacturers soon learned how to integrate everything on a single chip, sometimes laser-trimming components during manufacture, and selling a complete solution in a single DIP package. Analog design was forever changed.

Still, if someone is just starting out learning electronics, it is good to know that transistors exist for something else besides no-moving-parts switches. I still think the best way to learn about them is through bench prototyping and carefully maintained notebooks documenting your work. You can read all the theory you think you need, and execute innumerable SPICE models on your PC, but there is just no substitute for learning with real parts driven by real voltages and generating real currents. Well, real except when you get involved with reactances… but that’s another story.
 

Ratch

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The object of determining the device characteristic is to control the amount of base current. If the base resistor is large enough, the base will behave as though the base circuit is open. How large do you think the base resistor needs to be before the BJT can't tell the difference between the resistor and an open circuit? A good current source will have a resistance that large.

A BJT cannot tell what is connected to it. In its active region, it always acts independently as a transconductance amplifier. If you drive the BJT with a current source, then you have a circuit consisting of a BJT and a external resistor, which is not the same as a BJT alone. The BJT does not respond to a current source, but instead to the voltage produced by the current source.

Ratch
 

Laplace

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The BJT does not respond to a current source, but instead to the voltage produced by the current source.
Yes, that was my whole point. It is the base current that controls Vbe. The current source provides whatever voltage is necessary for the base to accept that amount of current. That is why the base is biased with current.
 

Ratch

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Yes, that was my whole point. It is the base current that controls Vbe. The current source provides whatever voltage is necessary for the base to accept that amount of current. That is why the base is biased with current.

Again, you are talking about the circuit consisting of an external resistor and a BJT. You would not be able to say the base current is controlling the Vbe unless the BJT collector current was controlled by the Vbe. The fact that the BJT is a transconductance amplifier enables the Vbe to adjust itself to conform to whatever current exists in the collector. It is not Ib that is controlling Ic, it is the Vbe caused by the voltage necessary to make Ib happen.

Ratch
 

Laplace

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The fact that the BJT is a transconductance amplifier enables the Vbe to adjust itself to conform to whatever current exists in the collector.
That would be something really interesting to see! What sort of test circuit would be necessary to demonstrate that collector current controls Vbe?
 

Ratch

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That would be something really interesting to see! What sort of test circuit would be necessary to demonstrate that collector current controls Vbe?

I never said the collector current controls Vbe. In fact, I said many times the Vbe controls the Ic. But if you adjust Ic by applying a voltage across Re through the base-emitter, naturally Vbe is going to adjust to a value in accordance with the BJT transconductance present at that current value. It has to do so. That just proves that the Vbe controls Ic in the active region. Also the physics of the BJT show it to be true.

Ratch
 
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