Question about BJT voltage divider biasing method

Thread Starter

simpsonss

Joined Jul 8, 2008
173
hi all,

Sorry for mistake at the title. it is a base biasing but not a voltage divider biasing.

Basically i'm trying to understand the operation of a BJT (2N3904). I had tried on plotting out the Ic versus Vce curve. What i get from the plot is that when i fixed Ib= 0.02mA Ic is almost constant when Vce is more than 0.6V. ok this make sense.

Then, i go for base biasing method. Rc i put a 3Kohm and Rb i put a 3.9Kohm resistor. While for Vdd i supply in a 5Vdc. When i tried to measure Id, Ic and Vce i get this results. Ic=1.7mA and Ib=1.10mA and Vce = 0.02V. So it means my BJT is not biased(am i right?:confused:). How can i calculate the resistor value to bias 2N3904?

thank you.
 

Jony130

Joined Feb 17, 2009
5,024
Your BJT is in saturation region.
You need Rb>>200*Rc to be in active region.
For example we wont Ic = 1mA and Vce= 2.5V
Rc = 2.5V/1mA = 2.5K = 2.2K
And form datasheet we get "beta" (Hfe) Hfe = Ic/Ib = 230 (current gain)
http://www.fairchildsemi.com/ds/2N/2N3904.pdf
Figure 1 "Typical Pulsed Current Gainvs Collector Current"
So Ib = 1mA/230 = 4.3uA
Rb = (5V - 0.6V)/4.3uA = 1MΩ

And this type of BJT biasing is good only for saturation.
Becaues in active region hfe is Ic depending and virtually every single BJT will be have different current gain

http://forum.allaboutcircuits.com/showthread.php?p=234782#post234782
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
And this type of BJT biasing is good only for saturation.
Becaues in active region hfe is Ic depending and virtually every single BJT will be have different current gain
So do u mean that base biasing method is good only for saturation region but no for active region? i cant get what u mean because i thought that normally we bias a transistor to an active region so that it operate properly. Can u explain a bit more about your statement?

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Another question is that if my BJT is in saturation region, i'm not able to calculate the Ic value by using this formula.

Ic = β ((Vcc -Vbe)/Rb)
because if Ic/Ib , β = 1.5 which is not the value stated in the datasheet as u mentioned.

thanks again for the replies.
 
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Jony130

Joined Feb 17, 2009
5,024
So do u mean that base biasing method is good only for saturation region but no for active region? i cant get what u mean because i thought that normally we bias a transistor to an active region so that it operate properly. Can u explain a bit more about your statement?
Well of-course you can use "base biasing" in your circuit.
But this is not a "good" circuit.
Becaues the beta of a BJT vary with collector current and temperatures.
In addition every single BJT will have different current gain.
So you'll have to choose RB resistor individually each time when you change the BJT.
So in practice we use this method:
http://1.bp.blogspot.com/_sqidiQSxJp0/S_v5LyUsimI/AAAAAAAAAAo/SZGUomHMrkM/s1600/Imagen1.jpg

Another question is that if my BJT is in saturation region, i'm not able to calculate the Ic value by using this formula.
Ic = β ((Vcc -Vbe)/Rb)
because if Ic/Ib , β = 1.5 which is not the value stated in the datasheet as u mentioned.
thanks again for the replies.
Becaues in saturation Ic = β * Ic don't hold anymore.
Collector current cannot be greater than:
Vcc/Rc = 5V/3K = 1.6mA.
And if base current is equal
Ib = (5V - 0.65V)/3.9K = 1.1mA
So if β=100
Ic must be large then:
Ic = 100* 1.1mA = 110mA
Voltage drop on Rc resistor
VRc = 110mA * 3K = 330V
So this voltage is impassible in this circuit as VRc_max = 5V
So BJT can not force this amount of current trough the load (ohms law and Kirchhoff law must hold ).
The only think that BJT can do in this situation is to full open.
And when transistor is full on (saturated) then he act just like a short circuit (switch)
http://forum.allaboutcircuits.com/showthread.php?p=165135#post165135
 
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