Why does a BJT enter saturation?

Question

I'm a beginner in EE, learning about the forward active mode of transistors. The equation, Ic = Ib * BETA, and Ic=Ie * ALHPA, are referred to as the ways to figure out what base resistor you should have when running forward active mode. What is Ic and Ie? Do these mean the amount it lets in, or some ratio?

Along with this, how do I use these numbers? On this site, https://learn.sparkfun.com/tutorials/transistors they give an example where they say Ic is 100mA. From my understanding, that means that if it hooked straight up to the positive and negative terminals it passes 100mA. But what if there are resistors in series? What effect does this have?

Thank you very much. I'm sorry if there is a term for this I don't know, by all means give me something to google.

-edits-

Thank you for the name suggestion.

Answer 1

The equation, Ic = Ib * BETA, and Ic=Ie * ALHPA, are referred to as the ways to figure out what base resistor you should have when running forward active mode. What is Ic and Ie?

\$I_c\$ is the collector current. That is, the amount of charge flowing in to the collector per unit of time.

\$I_e\$ is the emitter current. In a circuit theory class, this will be the amount of current flowing in to the emitter terminal. But in every-day usage, we might define it to be the current flowing out of the emitter terminal as that is more likely to be a positive number. A good source will provide a schematic with arrows indicated the direction to be considered positive for every current they want to talk about.

Do these mean the amount it lets in, or some ratio?

\$I_c\$ and \$I_e\$ are currents. \$\beta\$ and \$\alpha\$ are ratios.

Usually in EE a variable named \$I\$ will be a current.

When talking about BJT's, the important ratio is \$I_c/I_b\$. This is often designated by \$\beta\$ or \$h_{fe}\$. (\$\alpha\$ might sometimes also be important but it's much more rare for it to come up in a practical circuit solution)

On this site, https://learn.sparkfun.com/tutorials/transistors they give an example where they say Ic is 100mA. From my understanding, that means that if it hooked straight up to the positive and negative terminals it passes 100mA.

If they say the current is 100 mA, they mean the current is 100 mA in whatever circuit they're talking about. If that circuit has the collector connected directly to the power supply, that's what they mean. If they're talking about a circuit with resistors, then they mean the current flowing in to the collector in that circuit.

But what if there are resistors in series? What effect does this have?

If there's a resistor (and nothing else) between the power supply and the collector, then any current into the collector will have to flow through that resistor to get to the transistor. Ohm's law will tell you what the voltage drop must be across the resistor for that current to flow.

Answer 2

The main point to takeaway is the relation \$ I_C = \beta I_B\$ is a model of how a BJT behaves when operating in forward-active mode. As an extreme example, one cannot simply apply a negative current to the base of an NPN BJT and expect negative \$\beta I_B\$ current to flow in the collector.

The actual reason current flows in a BJT is extremely complicated and only described by modern solid state physics. However the models that have been developed to describe their behavior are in most cases "good enough" when applied within their regions of validity.

I have simplified your schematic slightly to illustrate the concept of saturation.

Take for example the following schematic,

Here the current into the base terminal is set by an ideal current source (only for convince in a Spice simulator) in a real circuit a resistor is often close enough to set the base current by ohms law.

The following figure is the simulations reults of a DC sweep of the current source injecting into the base terminal. The x-axis is base-current. The left y-axis is the collector voltage. The right y-axis is the collector-current.

We can see that when the collector-emitter voltage is greater than ~0.2V the forward-active relation holds true, namely, \$I_C = \beta I_B\$. As more base-current is applied more collector-current flows. However as more current flows more voltage is being dropped across R1 and the collector-emitter voltage is dropping. Eventually when the collector-emitter voltage becomes small enough the BJT is no longer in forward-active but in forward-saturation.

The textbook definition of forward-saturation for an NPN BJT is when Vbe is positive and when Vbc is also positive.

In saturation the BJT, loosely speaking it looks like a switch between collector and emitter. Slightly more complete model is a small mV level voltage source with low series resistance.

So when the NPN is behaving as a switch (saturated) it is the circuit around it that dictates what current will flow, in your case R1.