How can I fix a conceptual misunderstanding about BJT transistors in saturation?

I'll stick with the injection version of the DC (level 1) Ebers-Moll model. It was the first description made and it is more intuitive. So I imagine it may provide a better shot at it. (I will intentionally avoid the microelectronic view.)

injection model

Borrowing the image from this answer, here's the injection version drawn slightly differently:

The two diodes do not, by themselves, completely describe the bipolar transistor. The ultra-thin base region leads to some coupling between the junctions. (As you know, you can't make a bipolar transistor from two discrete diodes.)

Both diodes follow the Shockley diode equation. But they have different saturation current parameters (the dopant concentrations are lower in the collector and higher in the emitter.)

The coupling due to the ultra-thin base region is then modeled by two current-dependent current sources. Each junction diode has a different coupling coefficient: \$\alpha_{\small{F}}=\frac{\beta_{\small{F}}}{1+\beta_{\small{F}}}\$ and \$\alpha_{\small{R}}=\frac{\beta_{\small{R}}}{1+\beta_{\small{R}}}\$.

active mode

Let's start with the active mode case:

The BC junction is reverse-biased, above, and can be neglected. The resulting model is pretty simple. It's a diode and a current generator. The exact mechanism for the current generator is a matter of the microelectronics. But it is required in order to understand the active mode behavior (because of that ultra-thin base region.)

I think you already understand this one reasonably well. So let's drop it and move on.

saturation mode

Now both diodes are forward-biased per saturation mode requirements. Let's look at what happens by adding this new forward-biased BC diode, in parallel with the earlier forward-biased BE diode:

The full Shockley equation diode current, \$I_{\small{R}}\$, for the BC junction happens in the same way and for the same reasons as did \$I_{\small{F}}\$, before.

But the difference now is that \$I_{_\text{C}}\$ is reduced to \$I_{_\text{C}}=\alpha_{\small{F}}\:I_{\small{F}}-\underline{I_{\small{R}}}\$ and \$I_{_\text{E}}\$ is reduced to \$\left| I_{_\text{E}}\right|=I_{\small{F}}-\underline{\alpha_{\small{R}}\:I_{\small{R}}}\$, but \$I_{_\text{B}}\$ is increased to \$I_{_\text{B}}=\left(1-\alpha_{\small{F}}\right)I_{\small{F}}+\underline{\left(1-\alpha_{\small{R}}\right)I_{\small{R}}}\$.

That's the change that occurs.

summary

Let's make some quantitative predictions. That's always the best way to sort crap from useful knowledge.

I will need to know what LTspice uses for \$V_T\$ during a regular run. I can do this by diode-connecting a BJT, driving \$1\:\text{mA}\$ into it, and extracting the result using some values:

So \$V_T= 25.8649606\:\text{mV}\$.

In the above diagram you can see that I used a nearly ideal model in LTspice. But it has different values for the forward and reverse \$\beta\$ (so we can test what I'm saying.)

There's one added problem:

• LTspice uses a modified version of Gummel-Poon, not Ebers-Moll. The closest Ebers-Moll model is the hybrid-\$\pi\$, which only requires one saturation current, not two as does the injection version.

If you look above, you will see that my idealized bipolar model uses IS=10fA. The equivalent two saturation currents in the injection model are \$I_{_\text{ES}}=\frac{I_{_\text{S}}}{\alpha_{\small{F}}}\$ and \$I_{_\text{CS}}=\frac{I_{_\text{S}}}{\alpha_{\small{R}}}\$. I'll be using this information, shortly.

Some terms to define for the injection model use:

• \$\alpha_{\small{F}}=\frac1{1+\frac1{\beta_{\small{F}}}}\$
• \$\alpha_{\small{R}}=\frac1{1+\frac1{\beta_{\small{R}}}}\$
• \$I_{\small{F}}=\frac{I_{_\text{S}}}{\alpha_{\small{F}}}\left[\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)-1\right]\$
• \$I_{\small{R}}=\frac{I_{_\text{S}}}{\alpha_{\small{R}}}\left[\exp\left(\frac{V_{_\text{BE}}-V_{_\text{CE}}}{V_T}\right)-1\right]\$

From the injection model then also find:

• \$I_{_\text{C}}=\alpha_{\small{F}}\,I_{\small{F}}-I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[1-\frac1{\alpha_{\small{R}}}\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)\right]\$
• \$I_{_\text{B}}=\left(1-\alpha_{\small{F}}\right)I_{\small{F}}+\left(1-\alpha_{\small{R}}\right)I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[\frac1{\beta_{\small{F}}}+\frac1{\beta_{\small{R}}}\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)\right]\$
• \$I_{_\text{E}}=-I_{\small{F}}+\alpha_{\small{R}}\,I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)-\frac1{\alpha_{\small{F}}}\right]\$

In SymPy/Python:

VT = 25.8649606e-3                              # measured VT
ISAT = 10e-15                                   # model IS
def A(b): return 1/(1+1/b)                      # alpha factor
def IC(vbe,vce,vt,isat,br): return isat*exp(vbe/vt)*(1-exp(-vce/vt)/A(br))
def IE(vbe,vce,vt,isat,bf): return isat*exp(vbe/vt)*(exp(-vce/vt)-1/A(bf))
def IB(vbe,vce,vt,isat,bf,br): return isat*exp(vbe/vt)*(1/bf+exp(-vce/vt)/br)

I'm going to use \$V_{_\text{CE}}=100\:\text{mV}\$ and \$V_{_\text{BE}}=650\:\text{mV}\$ as a first test of the above injection model. I believe making \$V_{_\text{CB}}=550\:\text{mV}\$ will be forward-biased enough to matter enough when run also in LTspice, as validation of the above. I'll also use \$\beta_{\small{F}}=200\$ and \$\beta_{\small{R}}=20\$:

sci( IC(.65,.1,VT,ISAT,20), 4 )
'802.4051E-06'

The prediction is \$I_{_\text{C}}\approx 802.4051\:\mu\text{A}\$.

Now let's do an LTspice run:

Now, that is very close!!!

Let's try an extreme test by setting \$V_{_\text{CE}}=0\:\text{V}\$. In this case, the prediction is that the collector current will be negative. This is because the transistion from positive collector current to negative collector current (due to the differences in the two current generators) is at \$V_{_\text{CE}}=V_T\,\ln\left(1+\frac1{\beta_{\small{R}}}\right)=0.00126195567390989\approx 1.262\:\text{mV}\$. Below that value, expect a negative collector current:

sci( IC(.65,0,VT,ISAT,20), 4 )
'-41.0221E-06'

The prediction is in fact negative. It's \$I_{_\text{C}}\approx -41.0221\:\mu\text{A}\$.

Let's see what LTspice says:

Again, this is very close and the sign is right, as well.

Lastly, just to verify that the collector current reaches zero at \$V_{_\text{CE}}=1.26195567390989\:\text{mV}\$:

sci( IC(.65,0.00126195567390989,VT,ISAT,20), 4 )
'-182.1746E-21'

Which is as close to zero as one may wish to see.

LTspice calculates Ic(Q1) = -4.09828e-011.

So this is also a match.

There's another parlor trick, of sorts. Saturation can only be handled by the full 4-quadrant model, unlike active mode which only requires half of it.

Suppose \$V_{_\text{CC}}=3.3\:\text{V}\$ for an MCU and you need to ensure at or below \$V_{_\text{CE}}=100\:\text{mV}\$ with a collector load of \$160\:\text{mA}\$ that will be activated by an I/O pin and an NPN transistor. You can solve for the required \$V_{_\text{BE}}\$:

def VBE(ic,vce,vt,isat,br): return vt*ln(ic/isat/(1-exp(-vce/vt)/A(br)))
float( sci( VBE(0.16,0.1,VT,ISAT,20), 4 ) )
0.7869631         # required VBE
                  # note: bipolars are voltage-driven, not current driven
float( sci( IB(0.7869631,0.1,VT,ISAT,200,20), 4 ) )
0.0009892429      # required base current

(This would be \$\beta\approx 162\$.)

Remember, the expectation is no more than \$V_{_\text{CE}}=100\:\text{mV}\$, or \$\ge 3.2\:\text{V}\$ across the load at \$160\:\text{mA}\$. So the collector load can be treated as a \$\frac{3.2\:\text{V}}{160\:\text{mA}}=20\:\Omega\$ resistance.

Assume that the I/O pin will yield exactly \$3.3\:\text{V}\$ at \$0\:\text{mA}\$ and has a FET output resistance, when HI, of \$\approx 40\:\Omega\$.

Then it follows that the base resistor should be \$\frac{3.3\:\text{V}-786.9631\:\text{mV}}{989.2429\:\mu\text{A}}-40\:\Omega\approx 2500.364\:\Omega\$. Round that to \$2.5\:\text{k}\Omega\$ and the result should be very close.

I'll run both the exact and approximate values:

Again, quantitatively predictable.

Take note here. Normally, using a bipolar transistor as a switch (in deep saturation) isn't treated as predictable. Base current is usually just taken to be about \$\frac1{10}\$th of the collector current with a \$\beta\approx 10\$ assumption and the analysis stops there. It's easy.

But I want to point out that the full Ebers-Moll model is a powerful model and it just works right, even in saturation. It is only the full Ebers-Moll model that can handle saturation and active mode behavior (and reverse-active and cutoff modes, too), quantitatively. (It also, of course, clearly and precisely demonstrates that the bipolar transistor is a voltage-driven device. Not current-driven.)

It was and still remains a remarkable achievement.

What I've demonstrated is just how useful it is to consider the bipolar transistor as two customized diode models, laying one of them aside the other. And it works in all quadrants, as well.

The injection model was the first model to be disclosed in the literature because, in part, it's the easier model to use in our heads. It's just two simpler diode models (with current generators) laid side-by-side. So if you can put one of them into your head, it's easy enough to then just add the second one. And that is enough to work in all quadrants.

It's problems were:

• two model parameter saturation currents were required, one for each diode section, so simulation required two parameters where one would otherwise suffice (which would ease computation a bit)
• two current generators were required where one would otherwise suffice (which is easier to use in producing a small-signal model for analysis)

The first problem was solved by the fully equivalent transport model. The second problem was solved by the fully equivalent hybrid-\$\pi\$ model.

But the injection model better helps to see how it is that the forward BC diode current makes its way back. Please look it over a little more. Think over the ideas here.

They are all saying the same thing, just in different ways. But the original injection version is perhaps the easier way to see how saturation works.

I've mentioned these current generators due to the ultra-thin base region causing coupling in the bipolar transistor. I've demonstrated that these current generators properly predict the bipolar transistor behavior (at least, at DC and low frequency.)

You can either accept that idea and run with it or else you can demand to dig beyond the model and reach into the microelectroncs (physics) that is taking place inside the bipolar transistor and which can be summarized as "current generators" in Ebers-Moll's models, so well.

If you need that level of knowledge you can pick a textbook, such as any of the editions of:

• "Microelectronics" by Jacob Millman

Or you can read publications that came out early on -- on or before the mid-1960's -- such as the quite excellent volume used primarily for 3rd and 4th year undergraduate studies:

• "Physical Electronics and Circuit Models of Transistors" by Paul E. Gray (MIT), David DeWitt (IBM), A. R. Boothroyd (Queen's Uni of Belfast), James F. Gibbons (Stanford), 1964, vol. 2 of Semiconductor Electronics Education Committee (SEEC)

Obviously, there are the original papers to also read:

• "Physical Principles Involved in Transistor Action" by J. Bardeen and W. H. Brattain, April 1949
• "The Theory of p-n Junctions in Semiconductors and p-n Junction Transistors" by W. Shockley, 1949
• "Electrons and Holes in Semiconductors: With Applications to Transistor Electronics" by William Shockley, 1950
• "p-n Junction Transistors" by W. Shockley and G. K. Teal, 1951
• "Some Circuit Properties and Applications of n-p-n Transistors" by R. L. Wallace, Jr. and W. J. Pietenpol, 1951
• "Effects of Space-Charge Layer Widening in Junction Transistors" by J. M. Early, 1952
• "Large-Signal Behavior of Junction Transistors" by J. J. Ebers and J. L. Moll, 1954
• "The Dependence of Transistor Parameters on the Distribution of Base Layer Resistivity" by J. L. Moll, 1956
• "An Integral Charge Control Model of Bipolar Transistors" by H. K. Gummel and H. C. Poon, 1970
• "Direct Verification of the Ebers-Moll Reciprocity Condition" by B. L. Hart, 1971

I have copies (and have read them with a small bit of understanding) of all of the above papers and volumes, indicated above. I think the physics is best summarized in the Ebers-Moll model (or the Gummel-Poon model) and you can save yourself some reading by just accepting the ideas as granted.

(The MEXTRAM model is the most modern model. But I've not studied it in any detail. Nor have I read any of the original science publications related to it. So I don't have much to add about it.)

The two-diode model fails to properly describe the base-collector junction. It is at best misleading, because collector current flows the wrong way through it, something a regular diode would not permit. In isolation it behaves like a diode, that much is true, but during a transistor's normal operation, the B-C junction is clearly not behaving like a regular diode.

Also, if collector current in that model has to traverse the B-E junction, how is it possible to have \$V_{CE} < 0.7V\$?. And yet \$V_{CE} = 0.2V\$ is perfectly possible. This model fails in so many respects.

It's probably best, then, if you model the transistor in a way that better represents what's actually going on inside, in normal operation. Maybe you could picture it like this:

^{simulate this circuit – Schematic created using CircuitLab}

Here I use \$Z_{CE}\$ to represent a path for current directly from collector to emitter, bypassing the troublesome \$D_{BC}\$ and even \$D_{BE}\$ altogether. Its effective resistance (or conductance, however you prefer to interpret this element's role) has some complex relationship with \$V_{BE}\$. I'm not saying this is a good model, but I would claim that it better represents the various currents flowing than the misleading double-diode model.

I can't define the function \$Z_{CE} = f(V_{BE})\$ without a lot of hard work, but my point here is that thinking about the B-C junction as a just a diode is a gross over-simplification, and is only valid under very particular circumstances, none of which are present while \$V_{BE} > 0\$.

I can redraw my own model to represent a saturated transistor, with potentials labelled, arranged to have all higher potentials at the top, and conventional current always flowing downwards, which might make it easier to digest behaviour:

^{simulate this circuit}

There's no problem seeing here how emitter current \$I_E\$ will have contributions from three different current sources: positive \$I_3\$ through the forward biased \$D_{BE}\$, positive \$I_1\$ through the forward biased \$D_{BC}\$, and crucially, positive \$I_C\$ coming via the extremely low-impedance \$Z_{CE}\$. Verify for yourself that KCL is well observed:

$$ I_2 = I_C + I_1 $$

$$ I_E = I_B + I_C $$

This probably doesn't fully answer your question, but the question assumed the B-C junction is a diode, which is too naive as a model.

The problem is that the two-diode model does not explain what you want it to explain - i.e. a transistor in saturation.

What the two diode transistor model does explain is the internal structure of a BJT in a very simplified fashion.

It explains that there are two PN junctions than can be viewed as two diodes, and if you individually try to use each PN junctions, it will look like two individual diodes.

In reality the model is flawed because you already seem to know that two diodes bolted together do not make a transistor. The two PN junctions are just arranged so closely physically that pushing current through BE junction makes such conditions (electric fields, moving current carriers such as electrons and holes) that allow current to pass through from collector to emitter too, based on base current.

So the model does not even try to explain how a transistor works in saturation, or in active mode for that matter.

Which basically the model means that if you have a multimeter that has a "diode test" mode, you can quickly check the BE and BC junctions separately to see if the transistor should still be OK or damaged. It can definitely be used to check if an unkown BJT is of NPN or PNP type, depending on which way the internal "diodes" are. To a lesser extent, it should also be possible to determine the pinout, as base will have the common terminal, but the BE and BC junctions may have so small differences that it may or may not be possible to determine which way C and E pins are.

But here's the part that I need explained to me in really simple words, if possible, please! If that "diode" from base to collector is forward biased, how on earth does a large current \$I_C\$ get from collector to emitter? (See my blue arrow.) Isn't it going through the B-C "diode" in reverse?

Yes, it is crossing the forward biased base-collector junction in the reverse direction.

What confuses people are two things. One, based upon their experience with diodes they do not expect current to flow in the reverse direction at all, except for a small leakage current. Two, based upon their understanding of current flow, they do not expect (conventional) current to be flowing from the less positive collector to the more positive base (in a saturated NPN transistor).

Both of these expectations need to be changed to understand what is going on.

We will start with the second expectation, that (conventional) current always flows from the more positive to the less positive. That is simply not always true. There are two reasons why current flows in a circuit, drift and diffusion. Drift is current flow that is caused by an electric field (i.e. a gradient in potential. Diffusion is caused by a gradient in carrier density. Carriers will diffuse from regions of high concentration to regions of low concentration even against an electric field. Without an understanding of diffusion, and acceptance that it can work against an electric field, one will never arrive at a correct understanding of how semiconductors work.

Now, let's go back to how diodes work. At a PN junction, a built-in electric field is created. This is created by carriers diffusing across the junction. If you need a detailed explanation of how this built-in electric field comes about, ask, and it will be added to my answer. However, you may already understand, or simply accept that it is there. What is vitally important to know is that the electric field at a PN junction is in the reverse direction. By that I mean the electric field wants to push electrons from the P side of the junction to the N side, and wants to push holes from the N side of the junction to the P side. This defies many people's intuition about diodes, but is absolutely true. The reason why a reverse biased diode does not conduct a great deal of current is NOT that the PN junction prevents it, but rather because there are very few free electrons on the P side to flow across, and very few holes on the N side to flow across. But those holes that do exist on the N side, and those free electrons that do exist on the P side, will cross over the PN junction easily. They are said to be swept across by the built-in electric field.

To repeat, in a diode, there are few electrons on the P material to cross the PN junction. However, in an NPN transistor (with a forward biased base-emitter junction), there are many free electrons in the P-type base. And, whenever these free electrons arrive at the base-collector PN junction, they are swept across from the base to the collector. (Thus creating conventional current from the collector to the base.) If you need a detailed explanation of why there is a relatively high concentration of free electrons in the base, ask, and it will be provided.

Now, we are still left with the fact that the base (in a saturated NPN transistor) is more positive than the collector. The built-in electric field will want to push electrons across the base-collector junction from the base into the collector but won't the more positive base somehow "attract" the electrons back to the base? The answer is that some free electrons do cross the base-collector PN junction in the direction of collector to base. However, to cross the PN junction in this forward direction, they must diffuse across the built-in electric field. But the fact is, until \$V_{BC}\$ becomes very close to \$V_{BE}\$, the concentration of free electrons on the collector side is less than the concentration of free electrons on the base side, so that the concentration gradient will not favor electrons diffusing from collector to base. (That is, the concentration gradient will not favor conventional current from base to collector).

Addendum

From a comment:

Could you explain why base will have high concentrations of electrons in the base?

The emitter of a BJT is much more heavily doped than the base. 100 times more heavily doped is within the typical range. The more heavily doped a piece of semiconductor is, the more majority carriers (see note) there will be. Majority carriers are free electrons in N-type material, and holes in P-type material. As a result, in an NPN transistor, there will be many more free electrons in the emitter than holes in the base. The emitter current consists of two components: electrons traveling from the emitter to base, and holes traveling from the base to the emitter. Because the base has few holes, the majority of the emitter current consists of free electrons traveling from the emitter to the base.When current passes across the emitter-base junction, it is therefore mostly majority carriers from the emitter passing into the base.

Free electrons and holes tend to recombine. However, in the base there are few holes, so that the free electrons which crossed into the base tend to "live" a long time in the base without recombining. The recombination process in the base is "slow" (compared to other processes taking place). Also, the base is thin. So, it is likely that free electrons in the base will reach the base-collector junction before they recombine. As a result, most of the free electrons in the base of an NPN transistor pass right through the base and reach the collector. To give an idea of how many free electrons which begin in the emitter pass right through the base and reach the collector, we consider two transistor parameters, \$\alpha\$ and \$\beta\$. Without straying too far from the truth, the fraction that succeed in this trip is the parameter \$\alpha\$ and is very close to 1 in most transistors. Again, without straying too far from the truth, the ratio between the number of free electrons that succeed in reaching the collector to the number that fail is the parameter \$\beta\$, and can range anywhere from about 20 in power transistor to about 400 in a small signal transistor.

Short answer:

There are 4 mechanisms for significant low-frequency currents in a BJT:

1. Forward conduction of emitter diode.
2. Transistor effect.
3. Forward conduction of collector diode.
4. Reverse transistor effect.

In the active region the collector diode is reverse biased, so both 3 and 4 are negligible.

When the transistor is driven into saturation, as 2 through the load causes v_C to approach v_E , the collector diode becomes marginally forward biased while the emitter diode is decidedly forward biased. Current 1 can be increased further, but the corresponding increase in 2 gets balanced out by 3 and 4 because the collector diode forward bias is slightly increased by 2 through the load.