I am not getting as much time as I would like to study this stuff. Thats the reason for these late replies.

For Vcc = 10V; β = 100 and Rc = 100Ω
To saturate this transistor the base current must be larger than
Ib > = (Vcc/Rc)/β = 100mA/100 = 1mA
If we increase Rc resistor to say 500Ω and Ib remains unchanged (1mA).
The transistor will remain in saturation region. The only think that will change is Ic and Ie.
Ic ≈ 10/500Ω ≈ 20mA and Ie ≈ Ib + Ic = 21mA.
And transistor will start leaving the saturation region if the base current is lest than Ib < 20mA/β.

In reference to this: Increasing the resistor value means that less current (200uA) is needed to put the transistor in saturation?

And is it true that with such a small current needed to hold the transistor in saturation, that the transistor in my circuit will most probably be in saturation?

 

If we use Ib > Ic/(βmin * K ) the transistor will be in saturation?
Shouldn't K be less then zero?
Let's consider this example.

Assuming that the transistor has minimum beta βmin = 100.
The condition for transistor to be in saturation without using forced beta:
Ib > (Vcc/Rc)/βmin = (10V/1K)/100 = 10uA. (1)
If we use the forced beta with driving factor K = 5.
Ib = Ic/(βmin * K ) = (10V/1K)/(100*5) = 2uA. (2)

To make sure the transistor is in saturation, should we use (1) instead of (2)?

I'm embarrassed because once again I made a stupid mistake.
For me it is obvious that the overdrive factor should increase Ib.
The correct formula should looks like this:
Ib > (K * Ic)/βmin

 

In reference to this: Increasing the resistor value means that less current (200uA) is needed to put the transistor in saturation?

Yes.

And is it true that with such a small current needed to hold the transistor in saturation, that the transistor in my circuit will most probably be in saturation?

I don't now because you don't tell as what Ic current you need.
And pleas read post 19. And always use this Ic/Ib = 10 (Forced Beta).

 

The correct formula should looks like this:
Ib > (K * Ic)/βmin

Thanks. That makes sense now.
I think it would be great if you also edit the mistake in that old thread so that no one has to be confused.

 

Yes.


I don't now because you don't tell as what Ic current you need.
And pleas read post 19. And always use this Ic/Ib = 10 (Forced Beta).

Ok I will reword this. What I meant to ask was more along the lines of....

If saturation occurs at such a small current, then driving an LED or a motor which requires probably over 10mA, I will probably be putting the transistor into saturation mode?

I know this may not necessarily be true, but I am unable to try all the permutations of Vcc and R.

From recent readings and I am surprised that I have only just realised, actually I am not that surprised at all, is that when using the transistor as a switch, then it should be in saturation mode.

is this correct?

and also if its in saturation mode then it uses forced beta, which means that its not offering much amplification.

is that correct?

Don't roll your eyes....

 

From recent readings and I am surprised that I have only just realised, actually I am not that surprised at all, is that when using the transistor as a switch, then it should be in saturation mode.
is this correct?

Yes

and also if its in saturation mode then it uses forced beta, which means that its not offering much amplification.

is that correct?

Yes.

 

Looking back at these posts I am not sure I have a handle on this Vce_sat value that I mention here getting from datasheet.

Ok

anyway...

I then have to look to datasheet to get Vce_sat ≈ .003 V
from "Forced Beta"(europe) : Ic/Ib = 20
(you mentioned this in another post)

I (think) used figure 11.
http://www.onsemi.com/pub_link/Collateral/P2N2222A-D.PDF

and used the bottom line. But I then think I cross referenced the Ic @ 20(Beta Europe)

For european BJT Ib/Ib = 20

on the bottom axis, to get roughly 0.003

I don't think this is what I should have done to get the Vce_sat value.

Can anyone please clarify?

Thanks

 

To understand how BJT work, first you need to understand how a constant current source work. Are you familiar with the concept of a current source ?
The BJT's in active region act just like a base current controlled collector current source.
See this example show a BJT in active region.

If base current is flowing the BJT is ON and collector current is BETA (β or Hfe) times larger than the base current (Ic = β*Ib).

And Ic = β *Ib and Ie = Ib + Ic is a basic principle of a transistor "action".

Note that arrow on a BJT (on a emitter) symbol show how current will be flow through BJT. Of course from "+" to "-".

The BJT work very similarly to the tap water valve.
A water valve is always used as a device to control the flow of water. Similarly, always think of a bipolar transistor as a device used to control electric current flow by assistance of a base current.
If base current (Ib) flows hen BJT is ON so there must flow β times Ib current flow through collector. See this picture
http://images.elektroda.net/94_1250754403.png




You are doing a good job with you answers. And yes 10V is constant because this is the power supply voltage.

In our first example we forced Ib current to be equal to 10μA

So Ic = β * Ib = 1mA and voltage across Rc:
VRc = Ic * Rc = 1V so the Vce = Vcc - Vce = 1V (II Kirchhoff law).

As you can see in first two examples BJT work in active region and Ic current is equal to: Ic = β*Ib = 5mA

In the second example (Ib = 100μA), the transistor is on the edge, between the saturation and active region. Ideally this saturation voltage would be Vce = 0V but in real life it is impossible to happen.

In the last exampleIc = β*Ib = Ic = 100*1mA = 100mA don't holds anymore, do you know why?
Because now we have Rc in the circuit so Ohm's and Kirchoff's law must hold also.
What is the max Ic that can flow in this circuit? We all knows the Ohms law, so Ic_max = Vcc/Rc ≈ 10mA.
BJT tries to create a situation in which the collector current Ic = β*Ib = 100mA. And he lowers his collector-emitter voltage to Vcs_sat voltage. The BJT in full ON. Transistor is in saturation region. And in saturation Ic = β*Ib don't hold anymore. And in saturation the collector is equal to:
Ic_sat = (Vcc - Vce_sat)/Rc ≈ Vcc/Rc

And emitter current is always equal to Ib + Ic = 1mA + 10mA = 11mA

Here you have a more detailed explanation about saturation region.

I hope that know you see that saturation depends only on the Rc value, Vcc and transistor β.

As a small test try to find Ib needed to saturated this transistor in this two circuit.

In first we have β=100, and in the second one we have β=200

Should Ib neglected in active region...??

 

Should Ib neglected in active region...??

There is no simple answer for this question. Its all depend on a specific circuit. Most of a time we can ignore the base current at first glance.
For example: for you current source circuit we can ignore Ib because Ve > Vbe, so by knowing Ve we immediately can find Ie = Ve/Re