How do BJT transistors work in a saturated state?

Question

This is what I know about NPN BJTs (Bipolar Junction Transistors):

• The Base-Emitter current is amplified HFE times at Collector-Emitter, so that Ice = Ibe * HFE
• Vbe is the voltage between Base-Emitter, and, like any diode, is usually around 0,65V. I don't remember about Vec, though.
• If Vbe is lower than the minimum threshold, then the transistor is open and no current passes through any of its contacts. (okay, maybe a few µA of leak current, but that's not relevant)

But I still have some questions:

• How the transistor works when it is saturated?
• Is it possible to have the transistor in open state, under some condition other than having Vbe lower than the threshold?

In addition, feel free to point (in answers) any mistakes I made in this question.

Related question:

• I don't care how a transistor works, how do I get one to work?

Answer 1

Saturation simply means that an increase in base current results in no (or very little) increase in collector current.

Saturation occurs when both the B-E and C-B junctions are forward biased, it's the low-resistance "On" state of the device. The properties of the transistor in all modes, including saturation, can be predicted from the Ebers-Moll model.

Answer 2

Your \$I_{CE}\$ = \$I_{BE} \times h_{FE}\$ isn't quite right. This equation shows what the collector current could be if given sufficient collector voltage. Saturation happens when you don't give it enough voltage. Therefore, in saturation, \$I_{CE} \lt I_{BE} \times h_{FE}\$. Or you could look at it the other way around, which is that you are supplying more base current than needed to handle all the collector current the circuit can provide. Put mathematically, that is \$I_{BE} \gt I_{CE} \mathbin{/} h_{FE}\$.

Since the collector of a NPN will act like a current sink and in saturation the external circuit isn't giving it as much current as it could pass, the collector voltage will go as low as it can. A saturated transistor typically has around 200mV C-E, but that also can vary a lot by the design of the transistor and the current.

One artifact of saturation is that the transistor will be slow to turn off. There are extra "unused" charges in the base that take a little while to drain out. That's not very scientific and only roughly described the semiconductor physics, but it's a good enough model to keep in your mind as a first order explanation.

One interesting thing is that the collector of a saturated transistor is actually below the base voltage. This is used to advantage in Schottky logic. A Schottky diode is integrated into the transistor from base to collector. When the collector gets low when it's nearly in saturation, it steals base current which keeps the transistor just at the edge of saturation. The on state voltage will be a little higher since the transistor isn't fully saturated. The advantage is that it makes the off transition faster since the transistor is in the "linear" region instead of in saturation.

Answer 3

1. When it's saturated, the collector current is not \$h_{FE}\$ times the base current anymore. It's less, how much, it depends on the rest of the circuit (I'm talking about the simplest model you can think of). In saturation, the \$V_{CE}\$ voltage can be considered more or less constant and you can call it \$V_{CEsat}\$, let's say around \$0.2\mathrm V\$. TYour BJT is saturated when both its BE and BC junctions are active. That limits the \$I_C\$ current to less than \$I_B h_{FE}\$ and pins the \$V_{CE}\$ voltage drop to \$V_{CEsat}\$.

2. Why do you care of having your BJT in open state if there's no current going through it? It's like having your tap open with no water in the pipe :D

Answer 4

The emitter resistance connected means the transistor will go to saturation , but the base resistance and collector resistance will remain same .Batter you draw a circuit and calculate base current ,then you will good result.