How to bias a transistor using emitter bias method?

Thread Starter

simpsonss

Joined Jul 8, 2008
173
hi,

I would like to learn about the emitter bias biasing. From my reference book, the circuit is as below.

untitled3.JPG

But i'm having problem in identifying the 3 resistors value. How am i gonna put in the value? what value to choose?

thank you.
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
To pick components you actually need some targets in mind.

Gain?
Signal swing?
Output / input resistance?
ok.if i would like to have a 10 times in gain.
signal swing means what? the Amplitude?
Another question is why i need to know the output or input resistance?

I'm a newbie in design BJT biasing.
 

shteii01

Joined Feb 19, 2010
4,644
I think I found example that have the same circuit. The key to the whole thing is to have transistor in forward-active mode, that is Vce>Vbe(on).

Pull up the data sheet for that transistor and look up the Vbe(on). Then put voltmeters into the circuit to measure Vbe and Vce. Start playing with resistors until you get Vce > Vbe while Vbe=Vbe(on).

Also, the book says that the rule of thumb is that amplification factor is about 50 to 300. So gain of 10 might be too low to experiment with.
 

shteii01

Joined Feb 19, 2010
4,644
From my datasheet, i can see Vbe(sat) but no Vbe(on). So is this the same?

thank you.
I am going to take the coward's way out. Assume Vbe(on) is between 0.6 and 0.7 volts. Now start playing with resistors until you get Vbe equal to 0.7 volts, and Vce greater than 0.7 volts.
 

Wendy

Joined Mar 24, 2008
23,408
Hint: base emitter to -5VDC resistance is β (transistor beta) X R3. There is no Base to Collecter resistance. You should now be able to establish the voltage on the base, and work it from there.

2nd Hint: Ic + Ib = Ie
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
can i said that when Vce is equal to Vcc means my transistor is in cutoff mode?

So for currently i vary my Rc value. And i get Vbe=0.66 and Vce = 2.45V. So could i say that i'm working in the correct result?

thanks.
 

shteii01

Joined Feb 19, 2010
4,644
So for currently i vary my Rc value. And i get Vbe=0.66 and Vce = 2.45V. So could i say that i'm working in the correct result?

thanks.
That looks good. You are in forward-active mode, which is where you have nice linear relationship Ic=β(Ib) for current gain.
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
That looks good. You are in forward-active mode, which is where you have nice linear relationship Ic=β(Ib) for current gain.
How do i know whether it is in active mode? depends on the value for my Vce? What if my Vce = Vcc? Does it still in active mode?

thanks.
 

shteii01

Joined Feb 19, 2010
4,644
How do i know whether it is in active mode? depends on the value for my Vce? What if my Vce = Vcc? Does it still in active mode?

thanks.
I can give you the definition from the book.
Forward-active mode:
Vbe > 0
Vbc < 0

You need to meet both of these conditions. Stick the voltmeters on that circuit and see what you got. You already said that Vbe=0.66 volts, so first condition is met. What do you have for Vbc?
 

Wendy

Joined Mar 24, 2008
23,408
Does not make any sense.
Perhaps because you have dismissed it without studying it? I will explain in my next post.

In a common collector amplifier (which relates) the resistance as measure from base to ground (or in this case, V2 (-5VDC)) is the formula I stated.
 
Last edited:

Wendy

Joined Mar 24, 2008
23,408
your R3 means Rb or Rc on my circuit?

thanks.


I meant the component designation on your schematic, which is not labeled Re, Rc, or Rb, but is labeled R1, R2, and R3 (though I understood what you meant). Component designations are your friend, always.

If the β of the transistor is 100, you will have 100KΩ from the base of the transistor and through R3. The battery is in parallel with this resistance, but the resistance is real and can be used to determine the voltage you will see at the base and across R2.

The currents in this circuit (true for almost all transistor circuits) is

IR3 = IR1 + IR2, or Ie = Ib + Ic

In general for these designs, Ic ≈ Ie

You provided the schematic, so I am working with that.

One last Hint, the collector current and voltage is almost entirely dependent on the emitter current, unless the transistor saturates.

.
 
Last edited:

Jony130

Joined Feb 17, 2009
5,487
For voltage gain 10V/V
Rc/Re = 10
So for Rc = 1KΩ ---> Re = 100Ω
Vce = 0.5 * V1+V2 = 5V
So
Ic = 5V/1.1K = 4.5mA

Rb = ( V2 - Vbe - Ve)/Ib

Ve = 4.5mA * 100Ω = 0.45V
Vbe = 0.7V
Ib = Ic/β = 4.5mA/160 = 28μA

Rb = (5V - 0.7V - 0.45V)/ 28μA = 3.85V/28uA = 137K = 140KΩ

Another question is why i need to know the output or input resistance?
Read this post
http://forum.allaboutcircuits.com/showthread.php?p=153702#post153702
And remember Rin>> R_source and Rout<<Rload
 

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