NPN BJT saturation

For saturation of Vce at 0.2v; Does this mean that there is no further increase of Ic (Ie) even if i increase further Ib by increasing Vbe after this? Will the transistor itself be pushed beyond its limit, if i keep increasing Ib through Vbe, forgetting about the limit by external components?

Once it's saturated that's it, you gain nothing more by increasing base current.

So does it sound like this:

At cut off, which is from the range of 0 to 0.6 Vbe, the voltage drop across Vce is Vcc (as the transistor is like open switch)
As i increase Vbe or Ib, the voltage drop across Vce decrease until it reaches saturation at 0.2v.

If i were to plot a graph of Vce (Y axis) vs Vbe (X axis) , it should be a linear graph, at one end is (0.6, Vcc) and the low end is (Vbe-sat, 0.2)., am i right?

How do we determine by calculation what is the minimum required Vbe for the transistor to go into saturation mode?

Evelyn1234 said:

So does it sound like this:

At cut off, which is from the range of 0 to 0.6 Vbe, the voltage drop across Vce is Vcc (as the transistor is like open switch)
As i increase Vbe or Ib, the voltage drop across Vce decrease until it reaches saturation at 0.2v.

If i were to plot a graph of Vce (Y axis) vs Vbe (X axis) , it should be a linear graph, at one end is (0.6, Vcc) and the low end is (Vbe-sat, 0.2)., am i right?

Click to expand...

I wouldn't expect it to be linear?.

How do we determine by calculation what is the minimum required Vbe for the transistor to go into saturation mode?

Click to expand...

You probably don't, for a start bipolar transistors are current driven devices, so it's the base current that matters, not the voltage - I would suggest actually plotting the graph, and see what that tells you.

Here is the saturation curves for a high voltage 4A transistor.
You can see that at low collector current the curve has a square corner and at high current the corner is round.
If you increase to B current too much there is more heat in the transistor. 2A x 1V = 2 watts. Also the transistor will get very slow.


I picked a high voltage transistor to make a point. The B-E voltage is almost 1 volt. This is typical of a high voltage part. The silicon is different.

A reminder that Vbe is strongly temperature dependent: https://www.physicsforums.com/threads/vbe-temperature-coefficient-of-transistors.322302/

Evelyn1234 said:

.....................
How do we determine by calculation what is the minimum required Vbe for the transistor to go into saturation mode?

Click to expand...

Normally you don't concern yourself about the Vbe voltage at saturation, other then to use it when calculating the base resistor value to get the desired base current of about 1/10 of the collector current for full transistor saturation. Vbe is typically about 0.7V-0.8V under those conditions.

Notice that they said you never feed voltage to the base of a transistor (but to an emitter-follower is fine), instead you feed current.
A transistor has a range of Vbe. Some are lower and some are higher, even if the part number and manufacturer are the same. If a transistor is warm or hot its Vbe is less. Feed 0.7V to a hot little transistor base-emitter and it will be destroyed.
The hFE current gain of a transistor also has a range. Some are higher and some are lower, even if the part number and manufacturer are the same.
Some transistors saturate better than others, even if ....

Evelyn1234 said:

For saturation of Vce at 0.2v; Does this mean that there is no further increase of Ic (Ie) even if i increase further Ib by increasing Vbe after this?

Click to expand...

Even if you could get the Vce voltage down to the best minimum of zero volts, it would not matter too much because the 0.2 volt difference is probably insignificant compared to the voltage drop across the collector load resistance.

Will the transistor itself be pushed beyond its limit, if i keep increasing Ib through Vbe, forgetting about the limit by external components?

Click to expand...

Semiconductors have voltage and current limits across every terminal. Pursuse your spec sheet to see what they are.

Evelyn1234 said:

So does it sound like this:

At cut off, which is from the range of 0 to 0.6 Vbe, the voltage drop across Vce is Vcc (as the transistor is like open switch)
As i increase Vbe or Ib, the voltage drop across Vce decrease until it reaches saturation at 0.2v.

Click to expand...

In most circuits, the transistor and the collector resistor share the available Vcc voltage. You can define cutoff to be any value you want.

If i were to plot a graph of Vce (Y axis) vs Vbe (X axis) , it should be a linear graph, at one end is (0.6, Vcc) and the low end is (Vbe-sat, 0.2)., am i right?

Click to expand...

No, you are not. The relationship of Vbe to Ic is exponential. If you plotted Vce to Ib, then you would see a linear relationship.

How do we determine by calculation what is the minimum required Vbe for the transistor to go into saturation mode?

Click to expand...

First, determine the collector current to barely go into saturation. Then use Schlockey's equation to determine Vbe. That would be difficult to do because you do not know the needed parameters for your particular transistor. Best to use a emitter resistor and apply a voltage across it through the base to get the exact current you need. If necessary, you can directly measure Vbe to see what it is.

Ratch

Ratchit said:

The relationship of Vbe to Ic is exponential.

Click to expand...

That is why a transistor used to amplify an audio voltage signal without any negative feedback produces severe distortion at high levels (below clipping):