# Which method npn

Discussion in 'General Electronics Chat' started by duxbuz, May 24, 2014.

Feb 23, 2014
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2. ### crutschow Expert

Mar 14, 2008
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I'm not much into watching videos. Can you state what the differences are that you find confusing?

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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.

Firs assume the β = hfe = 100 is constant.

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

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4. ### duxbuz Thread Starter Member

Feb 23, 2014
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I presume saturation means saturated and cant take anymore

5. ### crutschow Expert

Mar 14, 2008
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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).

6. ### duxbuz Thread Starter Member

Feb 23, 2014
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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

Last edited: May 25, 2014
7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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

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• ###### 35c.png
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Last edited: May 25, 2014
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8. ### duxbuz Thread Starter Member

Feb 23, 2014
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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

9. ### duxbuz Thread Starter Member

Feb 23, 2014
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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

10. ### duxbuz Thread Starter Member

Feb 23, 2014
133
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On a second look I may have understood correctly

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, you understood correctly.

12. ### duxbuz Thread Starter Member

Feb 23, 2014
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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)

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?

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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 ?

14. ### duxbuz Thread Starter Member

Feb 23, 2014
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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

Last edited: May 27, 2014
15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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.

16. ### anhnha Well-Known Member

Apr 19, 2012
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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?

17. ### crutschow Expert

Mar 14, 2008
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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.

Apr 5, 2008
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19. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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.

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20. ### anhnha Well-Known Member

Apr 19, 2012
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Yes.
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.

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)?