Which method npn

Jony130

Joined Feb 17, 2009
5,487
The problem lies in your lack of knowledge about transistor.
Simply you don't know what saturation is, also in real life the transistor current gain is not constant but it's change with the collector current. And this is why different people use different techniques to find resistor values. And this is what confuse you.

And try answer this question:
Firs assume the β = hfe = 100 is constant.


Find Ic and Vce for Ib = 50μA; 100μA and 1mA.
 

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crutschow

Joined Mar 14, 2008
34,201
I presume saturation means saturated and cant take anymore
In general, yes. In the case of a BJT the definition is that both the base-emitter and collector-emitter junctions are forward biased. This means the transistor is fully on (as a switch) and the collector-emitter voltage is less than the base-emitter voltage. Typically to insure full saturation, a base current of at least 1/10th of the collector current is used (the transistor Beta value is ignored).
 

Thread Starter

duxbuz

Joined Feb 23, 2014
133
My results are:
5mA 5Vce
10mA 0Vce

I am not convinced!

And I am not posting the last result as it seems totally wrong!

Also i presume from this the 10V is constant and the resistance voltage changes?

But I am really almost guessing how to work this out!

Well i say guessing. I am guessing that I use Ohms Law to determine the voltage over the resistor which leave the Vce.

The other part I used Ic = Ib x β
 
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Jony130

Joined Feb 17, 2009
5,487
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


My results are:
5mA 5Vce
10mA 0Vce

I am not convinced!

And I am not posting the last result as it seems totally wrong!

Also i presume from this the 10V is constant and the resistance voltage changes?

But I am really almost guessing how to work this out!

Well i say guessing. I am guessing that I use Ohms Law to determine the voltage over the resistor which leave the Vce.

The other part I used Ic = Ib x β
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 think a reasonably simple yet useful description of the saturation process goes something like this:

Think of a transistor as a device that tries to create a situation in which the collector current, Ic, is β*Ib. The mechanism by which it does this is by controlling the "resistance" of the collector-emitter path through the device. If the present resistance is such that the collector current is less than β*Ib, then the transistor reduces the resistance. This generally results in a lower collector-emitter voltage which means that the voltage dropped across the external components increases which generally allows more collector current to flow. The reverse happens if the present collector current is more than β*Ib in order to reduce it.

But this process can only be carried so far. The transistor can only lower the collector-emitter voltage so far and, once that point is reached, the transistor no longer has any tools in its bag to increase the collector current up to the point it would like it to be, so the collector-emitter voltage stops dropping and the collector current is less β*Ib.

Ideally this saturation voltage would be 0V and the collector-emitter junction would simply look like a short circuit allowing any current to flow as long as it doesn't go above β*Ib (since, at that point, the transistor will come out of saturation and start increasing Vce in order to hold the current to β*Ib).

In practice, real transistors have a saturation voltage that is greater than 0V, but it is generally only a few hundred millivolts.

written by WBahn
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
 

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Thread Starter

duxbuz

Joined Feb 23, 2014
133
Thanks Jony I will digest this and get back to you. I have a busy day today so I will have to read it all later and respond. Many thanks
 

Thread Starter

duxbuz

Joined Feb 23, 2014
133
And I also think I have already spotted that I misunderstood the question in regards to providing the variables to give the result Ib

I was trying to use the Ib's that you provided to get the other variables.. will rework it
 

Thread Starter

duxbuz

Joined Feb 23, 2014
133
Ok

Firstly thanks Jonny for great tutoring on this subject. I have learnt a lot.

I finally got chance to read and digest the post today.

I understood the concept in the write up.

And now I understand why the last example you gave me came out at a strange result.


I only just read the last questions. And I found them tricky.

Being a novice some of the hurdles are probably really simple.

I can only seem to come up with the current of the circuit which is
10v/500 = 0.02A = 20mA

If I then do

.02 / Beta = 0.0002
0.0002 * Beta * R = 10V

I did this as I was unsure what I was doing!! so I see the result is 10V, which leaves 0V over the Vce. Hence the questions.


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)

If transistor in our circuit will work as a switch, you don't need Hfe from datasheet.
Most BJTs vendors define saturation region when Ic/Ib = 10 (called forced beta) .
And this is why we assume Hfe_sat = 10
http://www.onsemi.com/pub_link/Collateral/P2N2222A-D.PDF (figure 11).
And the most datasheet show Vce_sat for Ic/Ib = 10
So to be one hundred percent sure that your BJT will be in saturation you must use this so-called forced beta technique.
For european BJT Ib/Ib = 20

But if we design an amplifier, we use a minimum hfe value from the data sheet.

First problem:

Ic_sat = (Vcc - Vce_sat)/Rc

Ic_sat = (10V - 0.003V) / 500
Ic_sat= 0.019994
Ib = Ic_sat / Beta

0.0001999A
199uA

.0001999 x Beta x RC = V Rc
9.995 over Rc

19mA Ic


Second one:

Ic_sat = (10V - 0.003V) / 500
Ic_sat = 0.019994
Ib = Ic_sat / Beta

0.0000999A
99.9 uA

0.0000999A * Beta * RC = V Rc

9.99V over Rc
19mA Ic

How are these looking?
 

Jony130

Joined Feb 17, 2009
5,487
Your answer look good. But what is more important here, is do you understand why transistor will be in saturation if Ib current is larger than 200μA and 100μA?

Also are you able to find Ib current the get Vce = 5V ?
 

Thread Starter

duxbuz

Joined Feb 23, 2014
133
I can only work through and see that 100uA puts the transistor in saturation by working through, I cannot just see it through recognition.

Although now you have told me I might.

But if the resistor was increased in the circuit would this still put the transistor in saturation?
Ic_max = Vcc/Rc (I only quoted this because it was formula I haven't used before)

I have tried changing the resistor in the calculations it made no difference. My results always push the Amps over the Ic_max


As for the other question you just asked; I have already had an example from you using 50uA that gave me 5Vce

Ok I see a pattern forming with the resistance, the Ib's uA, the β and the V

60uA giving me 6V, using 500R giving me 3V all @ 100β
 
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Jony130

Joined Feb 17, 2009
5,487
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.
 

anhnha

Joined Apr 19, 2012
905
When we use a BJT as a switch (saturation / cut-off) we use forced beta Ic/IB = 10 --->Rb/Rc = 10 --- > Rb = 10 * Rc

Or Ib = Ic/(βmin * K )
where
K = 3...10 - overdrive factor
This is from an old thread: http://forum.allaboutcircuits.com/showthread.php?p=666864#post666864

I think I am confused now. Is it true that the larger Ib, the easier it is to make a transistor into saturation?
From the formula, K is only a limited range. Why not K < 10 or something else?
Am I missing something?
For example, if Ic/Ib = 10 (or 20) a transistor will be in saturation and if Ic/Ib < 10 (or 20) is the transistor still in saturation?
 

crutschow

Joined Mar 14, 2008
34,201
.......................
For example, if Ic/Ib = 10 (or 20) a transistor will be in saturation and if Ic/Ib < 10 (or 20) is the transistor still in saturation?
Any value of Ic/Ib \(\leq\)10 should put the transistor well into saturation. Once you reach saturation, any further increase in base current will have only a small effect on the Vce saturation voltage.
 

Jony130

Joined Feb 17, 2009
5,487
This is from an old thread: http://forum.allaboutcircuits.com/showthread.php?p=666864#post666864 I think I am confused now. Is it true that the larger Ib, the easier it is to make a transistor into saturation? From the formula, K is only a limited range. Why not K < 10 or something else? Am I missing something? For example, if Ic/Ib = 10 (or 20) a transistor will be in saturation and if Ic/Ib < 10 (or 20) is the transistor still in saturation?
I don't understand why you confused?



For this circuit with ideal transistor any base current large than:
Ib > (Vcc/Rc)/β our ideal transistor will be well into saturation.
Do you agree with this statement ?

But in real life ideal transistor don't exist. For any real world transistor the β is not constant. Beta varies with Ic, Vce, temperature. And what is worse, every single transistor will have different beta value and beta will changes for different operating conditions also.
Also in saturation Ic = Ib * β don't work.

To overcome this problem with beta and saturation we are forced to use " overdrive factor"/"Forced Beta" trick.
We simply increase the base current well beyond Ib > (Vcc/Rc)/β (beyond minimum beta). We do this to make sure that we have enough base current to put the transistor well into saturation for every condition we have in our circuit.
 

anhnha

Joined Apr 19, 2012
905
For this circuit with ideal transistor any base current large than:
Ib > (Vcc/Rc)/β our ideal transistor will be well into saturation.
Do you agree with this statement ?
Yes.
But in real life ideal transistor don't exist. For any real world transistor the β is not constant. Beta varies with Ic, Vce, temperature. And what is worse, every single transistor will have different beta value and beta will changes for different operating conditions also.
Also in saturation Ic = Ib * β don't work.
If so, can we use the formula above with minimum beta, βmin?
If we set Ib > (Vcc/Rc)/βmin will the transistor will be 100% in saturation?
I think there is not true. Otherwise, we don't need to use "Forced Beta" at all.

To overcome this problem with beta and saturation we are forced to use " overdrive factor"/"Forced Beta" trick.
We simply increase the base current well beyond Ib > (Vcc/Rc)/β (beyond minimum beta). We do this to make sure that we have enough base current to put the transistor well into saturation for every condition we have in our circuit.
Yes, I see that but I am confused about this formula.


Ib = Ic/(βmin * K )

where
K = 3...10 - overdrive factor
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)?
 
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