BJT current distribution

Question

I asked a similar question here a few days ago.

I'm asking this question here again because the answers were ambiguous.

Consider the first picture, no collector battery, only a base voltage is given. In this case the base-emitter junction would simply act as a normal diode, and hence the current \$i\$ would enter the emitter and leave through base, hence emitter and base current would be the same.

Next, let us add a battery to the collector as shown in the second picture. Now what would happen? I'm considering the following scenarios:

The same current \$i\$ (no additional emitter current due to the collector battery) would distribute to the collector in the ratio of beta. (Quite simple to imagine.)
After adding the battery a new current of amount \$L\$ would be pulled from the emitter so now the total emitter current is \$i+L\$. This new \$i+L\$ would be distributed in the ratio of beta to the collector.

Which one is correct? Is there any other possible scenario which I'm missing?

It depends on how much current the sources can supply, but it will probably fry your BJT. — Hearth, Dec 27 '20 at 18:12
@ the Photon... that question not however exactly the same, but they fundamentally linked, here is the link : https://electronics.stackexchange.com/questions/538383/emitter-current/538544#538544, and the set possibilities i can imagine right now is already written in my post, kindly see. — Sayan, Dec 27 '20 at 18:17
@Sayan In the future, don't put a space after the @ if you want us to get notified, I just saw this by chance. To elaborate, you should never hook up a BJT to a voltage source without some kind of current limiting element (usually a resistor) in series with the base-emitter junction, else you're going to fry it, and then there won't be any current going through it at all and it might be on fire. — Hearth, Dec 27 '20 at 18:36
@Hearth alright, assume the circuit is properly designed with the current limiting resistors, then what would happen out of these possibilities? I'm not interested here about the practical issue, i.e excess current, rather I want to understand the current distribution . — Sayan, Dec 27 '20 at 18:41
@Sayan Now it depends on how the resistors are configured, though. — Hearth, Dec 27 '20 at 18:43
@Hearth let's say the circuit is configured with emitter resistor, no collector resistor. — Sayan, Dec 27 '20 at 18:45
@Sayan The question posed here is fundamentally different from the one posed previously. Here, you are supplying a fixed voltage between the base and emitter. In the previous question, there is a fixed voltage across the base-emitter junction AND a resistor. Big, big difference. — Math Keeps Me Busy, Dec 27 '20 at 19:26
I think you are confused. Given the battery polarity, and assuming this is an NPN transistor, current would enter the base and leave through the emitter. If it is a PNP transistor, then the base emitter junction is under reverse bias and only a small leakage current would flow. — user57037, Dec 27 '20 at 21:57
@Sayan well, don't do that. Nobody else does it that way. It only creates confusion. — user57037, Dec 28 '20 at 10:13

score 3 · Answer 1 · answered Dec 27 '20 at 18:52

If we assume your voltage source is forcing a fixed voltage across the transistor that's low enough not to fry it, nothing you do to the collector (as long as you keep the c-b junction reverse-biased!) will change the base current, because the base current is wholly determined by the base-emitter voltage, just like any diode.

If you add current limiting resistors, what happens depends on where the resistors are. If there's a resistor in series with the base and none on the emitter (whether there's a collector resistor or not doesn't matter for this situation), nothing changes; the base-emitter voltage remains the same and thus so does the current, regardless of what's going on with the collector.

If there's a resistor in series with the emitter and not the base, then applying a collector current will increase the current through the emitter resistor, lowing \$V_{be}\$, reducing the base current. Again, the collector resistor doesn't matter as long as \$V_{cb} > 0\$.

If there's one in both positions, the base current will decrease when you apply a voltage to the collector, but in a more complicated way. The increased emitter resistor current will lower \$V_{be}\$, but the decreased base resistor current will raise it, and the overall effect will be a base current somewhere between the base-resistor-only case and the emitter-resistor-only case.

Math Keeps Me Busy · Answer 2 · 2020-12-28T01:22:59.583

The base current \$I_b\$ is pretty much determined by \$V_{be}\$ and the temperature of the base-emitter junction and the geometry of the transistor. \$I_b\$ is approximately exponentially related to \$V_{be}\$. It is relatively independent of \$V_{ce}\$ and \$I_c\$.

In the linear, or active region of the transistor, \$I_c\$ is pretty much determined by \$I_b\$, and they are more or less linearly related.

\$I_c \approx \beta I_b\$

In the active region, \$V_{ce}\$ has some influence on \$I_c\$ but it is limited. (That influence can be seen in the characteristic \$I_c\$ curves being slightly inclined.)

In the saturated region of the transistor, \$V_{ce}\$ is too low for the above relationship, and

\$I_c \lt \beta I_b\$

where \$\beta\$ is the constant that is used in the active region.

Here, \$I_c\$ depends largely on \$V_{ce}\$. In fact, when the transistor is fully saturated, \$I_c\$ depends almost exclusively on \$V_{ce}\$ and they are more or less linearly related.

Also, in the saturation region, \$I_b\$, while still mostly determined by \$V_{be}\$, becomes increasingly influenced by \$V_{bc}\$.

In saturation, Ib is determined not only by Vbe but also by Vbc. Some of the base current goes to the collector when operating in saturation. — user57037, Dec 27 '20 at 21:29
What I mean is that you ought to put a footnote on this statement: " It is relatively independent of Vce and Ic." The footnote should say "base current will increase if the transistor is in the saturation mode of operation" or some such. — user57037, Dec 27 '20 at 21:33

Sredni Vashtar · Answer 3 · 2020-12-28T10:05:52.853

As with your previous question, I believe you should approach this problem as a limit approach of increasing Rc. Moreover, you might gain some benefit from comparing two similar, but very different from the point of view of the base current, configurations.

A common emitter where current limiting is done via a base resistor, and an emitter degenerated configuration where current limiting is done by an emitter resistor. For the moment I'll just add the following picture

The second configuration is explained in more detail in the above linked question. In terms of load lines, you want to know what happens when you switch from a finite Rc (the slanted load line denoted by (2) on the graphs) to an open circuit (the horizontal load line denoted by (1) on the graphs, and highlighted by ticks).

As you can see the operating point goes from the active zone (I suppose the finite value of Rc was enough to keep the transistor out of saturation) to the point with Ic = 0 (and ideally Vce = 0 with a perfect transistor, let's not get into this level of detail). The big difference is that in one case Ib is a negligible contribution to Ie, while in the other Ib is nearly almost all of Ie.

If we go by the approximation Vbe=0.7V in active region, in the common emitter case with base resistor Rb we have that Ib is constant and equal to (Vbb - 0.7V)/Rb, while Ic and Ie change as Rc changes; in the other case with emitter resistor Re, it is the emitter current that remains constant at (Vbb - 0.7V)/Re while Ib and Ic change as Rc changes, as show in the graph in my other answer. (Vbb is the base battery voltage)

Finally, the reason you are confused is that by removing all resistances you have put yourself in a nearly impossible situation when you switch from Rc=0 to Rc=infinity.
If you keep a base current limiting resistor (either as Rb or as Re) then the above graphs and reasoning still apply, with the only alteration that the case Rc=0 will see the load line (2) as a vertical line.
But if you remove all limiting resistors then you will have a vertical load line on the base circuit as well, and you need a very precise Vbb to avoid exceeding the maximum base current limit. In theory, with an ideal transistor with an ideal base diode (vertical characteristic at Vbe, the one we used as an approximation to easily compute base current and emitter current in the two circuits above) any value higher than the exact value required will cause an infinite current and the destruction of the transistor. So, either you consider the exact nonlinear characteristics of the BJT (which, due to the steep nonlinearity will give little leeway before you kill the device), or you add limiting resistors and see what happens in the limit when their values go to zero (or to infinity in the case of Rc).

score 0 · Answer 4 · answered Dec 28 '20 at 02:56

You are missing the diffusion part of the bjt physics. Without anything connected to the collector the collector soon gets charged and starts to push off electrons if some happened to reach the collector. Base wire is the only way to a +voltage.

Diffusion has a substantial role here. Electrons which come from emitter have much larger thermal random motion than what's the thickness of the base. Only a small percentage of the electrons hit the base wire, most of them hit the big collector area and there they get sinked if there's some +voltage connected.

Ideally the collector battery shouldn't pull more current, it only makes a way for the electrons that base battery has pulled. When the collector battery makes a new way, more electrons can pass the base area because the base voltage rises less with certain emitter current (=less electron packing into the base region) This is why base current really controls collector current, CE voltage affects less if there's at least few hundred millivolts.

Early effect unfortunately partially spoils the Vce-independent Ic/ib control because the effective base thickness varies. See this: https://en.wikipedia.org/wiki/Early_effect

score 0 · Answer 5 · answered Dec 28 '20 at 10:28

Consider the first picture, no collector battery, only a base voltage is given. In this case the base-emitter junction would simply act as a normal diode, and hence the current i would enter the emitter and leave through base, hence emitter and base current would be the same.

Apart from the fact that you are using opposite current convention, this is correct.

Next, let us add a battery to the collector as shown in the second picture. Now what would happen?

If we assume the battery voltage on the collector is large enough to keep the base collector junction under reverse bias, then there would be no (or minimal) effect on the base current. The emitter current would be (1 + beta) * base current.

If the base collector junction comes under forward bias, then some current will flow from base to collector as well as base to emitter (and of course in that condition, collector to emitter also).

BJT current distribution

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