Collector current vs. collector-emitter voltage

Question

Why is the collector current in an NPN transistor in CE mode almost independent of the collector-emitter voltage?

Doesn't increasing the reverse bias voltage (Vce) decrease the width of the base (due to increase in width of depletion region) and as a result, aren't more electrons swept from the base to the collector region which in turn increases the collector current (Ic)?

Answer 1

There's a couple of clarifications I need to make before I can address your questions.

1. Although textbooks far too often show charts having \$V_{_\text{CE}}\$ as the x-axis for certain illustrations about collector current, they contribute to confusion when doing so as the models don't operate that way. Instead, it would probably be better to show reverse-biasing of \$V_{_\text{CB}}\$ as the x-axis. So I take exception to your phrasing, using \$V_{_\text{CE}}\$, and will instead address myself to reverse-biasing of \$V_{_\text{CB}}\$.
2. When the reverse-biasing of \$V_{_\text{CB}}\$ is small, less than 4 times \$V_T\$, what I'll write is no longer accurate because the classical p-n diode reverse conductance can no longer be neglected in the analysis. But as this conductance magnitude decreases by an order of magnitude for every \$60\:\text{mV}\$ increase in reverse-biasing, its importance quickly disappears and so can be neglected once the reverse-biased \$V_{_\text{CB}}\$ voltage difference exceeds 4 times \$V_T\$.

Why collector current in NPN transistor in CE mode is almost independent of the Collector-Emitter voltage?

Here, you need to go to W. Shockley's 1949 paper, "The Theory of p-n Junctions in Semiconductors and p-n Junction Transistors".

Under his assumption, that changes in collector and emitter barrier thicknesses did not affect the base-layer thickness, he was able to "prove" that the collector current was independent of the reverse-biasing of the collector with respect to the base. (Note that he explicitly mentions #2 above as a requirement for his proof.)

Doesn't increasing reverse bias voltage (Vce) decreases the width of the base (due to increase in width of depletion region) and as a result more electrons are swept from base to collector region and in turn increases the collector current (Ic)?

That would now be J.M. Early's 1952 paper, "Effects of Space-Charge Layer Widening in Junction Transistors".

(Here, Early assumes that the emitter-side of the barrier doesn't move -- which would await Gummel & Poon, 1970, "An Integral Charge Control Model of Bipolar Transistors" for a correction to that assumption.)

Early shows that as reverse biasing of \$V_{_\text{CB}}\$ increases, what he describes as a "barrier thickness" \$x_m\$ also increases and eats into the base thickness, thinning the base region.

A thinner base layer, he writes, has two effects:

• It "decreases recombination of injected minority carriers in the base layer since the average carrier diffuses across the narrower base in a shorter time" and therefore "increases the transport factor \$\beta\$, which is the fraction of emitted minority carriers which reach the collector."
• It leads to a "decrease in the impedance presented to minority-carrier current injected by the emitter." This is because the "impedance seen by this injected current depends on base-layer resistivity and base-layer thickness".

Answer 2

The power of functional explanations

Why is the collector current in an NPN transistor in CE mode almost independent of the collector-emitter voltage?

In addition to explaining this phenomenon through semiconductor theory (which would do a good job, for example, to take the exam in this subject) we also need a more general functional explanation to imagine what the transistor actually does when it keeps the current constant. In practice, it is what we use to understand circuits (as, for example, the driver does not calculate the trajectory of the car, but somehow intuitively turns the steering wheel as it should). So why dismiss it as "unscientific" when it is what we do our job with?

The advantage of the "functional explanation" is that it captures the most essential idea in the observed phenomenon and is therefore universal. For example, if we can explain here how a BJT keeps the current constant, then we can explain how a FET does it, or how tubes ever did.

Static resistance

Two centuries ago, Ohm found that...

V = 1 V, R = 1 kΩ, I = V/R = 1/1 = 1 mA: ... if we apply a voltage V (1 V) across a resistor with a resistance R (1 kΩ), a current I (1 mA) flows.

^{simulate this circuit – Schematic created using CircuitLab}

V = 2 V, R = 1 kΩ, I = V/R = 2/1 = 2 mA: If we double the voltage (2 V), the current also doubles (2 mA)...

^{simulate this circuit}

V = 4 V, R = 1 kΩ, I = V/R = 4/1 = 4 mA: ... and so on in the same way (4 V -> 4 mA)...

^{simulate this circuit}

As you can see in the graph below, when you sweep (change) the voltage from zero to 10 V, the current changes proportionally from zero to 10 mA.

So we perceive this resistance as something constant, static, primal... and use it to introduce the next type of resistance.

Dynamic resistance

Now let's imagine that you decided to reproduce Ohm's experiment now, two centuries later.

V = 1 V, R = 1 kΩ, I = V/R = 1 mA: For this purpose, you originally set the same voltage V (1 V) across a resistor with the same resistance R (1 kΩ) so the same current I (1 mA) flows. The only difference is that the resistor is variable... but you do not know that.

^{simulate this circuit}

V = 2 V, R = 2 kΩ, I = V/R = 2/2 = 1 mA: Now, when you double the voltage (2 V), I (unnoticed by you) also double the resistance so the current does not change (1 mA)...

^{simulate this circuit}

V = 4 V, R = 4 kΩ, I = V/R = 4/4 = 1 mA: ... and so on in the same way (4 V -> 4 kΩ -> 4 mA)...

^{simulate this circuit}

So you have the illusion that the resistance is infinitely high because the current does not change; the self-changing resistance creates the illusion of infinite resistance.

In the graph below, the voltage is swept from zero to 10 V at four values of the resistance - 1, 2, 4 and 8 kΩ. You can imagine that as the voltage increases, the resistor IV curve slopes to the right so that its intersection point with the V IV curve moves along a horizontal line.

The current (I) does not change because the numerator (V) and denominator (R) of the Ohm's ratio (V/R) change equally. So such a dynamic resistor acts as a constant current source.

Note that "dynamic resistance" is different from "differential resistance". It is a qualitative definition of this type of resistance and literally means "self-changing static resistance"; it serves to understand the phenomena. Differential resistance is a formal quantitative definition and represents a ratio of changes in voltage and current; it serves to calculate.

Conceptual current source

Passive current source

Now we can imagine what might be inside the circle of the current source symbol - a dynamic resistor.

^{simulate this circuit}

When the voltage changes from zero to 10 V, the internal resistance changes from zero to 10 kΩ so the current stays 1 mA.

Active current source

Usually (and here in CircuitLab) it is believed that there is also a voltage source inside. Therefore, we can only connect a resistor to such a "true current source".

^{simulate this circuit}

As you can see in the graph below, when you sweep (change) its resistance R from zero to 10 kΩ, the current does not change because the current source decreases its internal resistance (or increases its internal voltage).

Transistor current source

A transistor with constant input (base-emitter) voltage behaves as the dynamic resistor above. It does not contain a voltage source inside; so it is a passive current source. Let's investigate it. Open the Vref parameters window and carefully adjust Vref so that a 1 mA collector current flows (the base current vigorously changes because the base-emitter junction is another type - constant-voltage, of dynamic resistance).

^{simulate this circuit}

As you can see, it is not a perfect current source because of the Early effect.

Imperfect current sources

"Bad" conceptual current source

We can simulate this effect by connecting a resistor in parallel to the perfect conceptual current source.

^{simulate this circuit}

We see that the slope of the horizontal part increases.

"Bad" transistor current source

We can also connect a resistor in parallel to the collector-emitter part of the transistor...

^{simulate this circuit}

... to "increase" the Early effect.

Conclusions

• The BJT collector current is (almost) constant when the collector-emitter voltage changes because the transistor changes its collector-emitter static resistance in the same way.

• The transistor behaves as a current-stabilizing non-linear (dynamic) resistor.

• The dynamic resistance can be thought of as a "self-varying static resistance".

• It is different from the "differential resistance" which is a formal quantitative definition that serves to calculate, and represents a ratio of changes in voltage and current.

---

See also my Codidact paper.

Answer 3

This is the Early effect, named after Jim Early, one of the early workers in the field.

There is a compromise between current gain and magnitude of the Early effect, as increasing the base doping decreases the gain but also reduces the Early effect.

Answer 4

Early Effect simulation, Maple sheet, Syrup & share library.
Early voltage is (-) 5 V (test), generally it is > 50 V.
The curves are then more "horizontal".