Common-emitter amplifier question

Thread Starter

beijim

Joined May 11, 2012
3
First, sorry for my poor English.

The question is shown below.

diagram4c.jpg

Here are my calculations, is that right?

Ci) Rc:

Because Vce=Vcc/2 , so Vcc=Vce*2

Vcc=12V

Rc=(Vcc-Vce)/Ic

=(12-6)/15*10^-3

=400Ω

IB:

I know IB=Ic/βDC but i don't know how to find βDC :eek: !

Because I know RB, so I find another way :

RB=(Vcc-VBE)/IB

This time, I find that I don't know how to find VBE :confused: ...

Because I cannot find IB ...I also cannot find IE and
βDC .

Can anyone help me?

Finally, thanks for you help. :)
 

Jony130

Joined Feb 17, 2009
5,487
First of all your Vce must be equal to 6V and Vcc = 10V
As for the Vbe simply assume Vbe = 0.6V
And are you sure that R1 is 10K ohm ? If this is the case I'm very strange surprised.
Since most of a BJT as a current gain greater than 100.
And for condition as in you diagram the BJT will by in saturation.
 

MrChips

Joined Oct 2, 2009
30,618
You are told VCC = 10V.

Why do you think VCE= VCC/2 ?
This is a wrong assumption.

You are also told that R1 = 910KΩ

You do not need to make any assumptions about VBE. If this is small it will not affect the calculation for IB.

Assuming VBE= 0.6V will give a better estimate of IB.

 
Last edited:

Thread Starter

beijim

Joined May 11, 2012
3
Thanks for your reply. :D

I would like to know why Vce is 6V? (Sorry for my silly question, because I am a novice in Electricity. :p)

And here are my corrections, is that right?

Ci) Rc:

Given : Vcc=10V, Vce=6V, Ic=15mA

Rc=(Vcc-Vce)/Ic

=(10-6)/15*10^-3

=266.67Ω

Ib:

Rc=(Vcc-Vbe)/Ib

910k=(10-0.6)/Ib

Ib=10.33μA

Cii) Ie:

Ie=Ic+Ib

=15m+10.33μ

=15.01mA

βdc:

βdc=Ic+Ib

=15m/10.33μ

=1452.08
 
Last edited:

Audioguru

Joined Dec 20, 2007
11,248
A transistor is NEVER biased like that because Beta is a range of numbers. Its beta might be high or it might be low and it changes when the temperature changes.

Usually the base of a common-emitter amplifier transistor is biased from a voltage divider and the transistor has an emitter resistor.
 

mlog

Joined Feb 11, 2012
276
Chips and Martin pretty much answered the question. The assumption is that the silicon transistor begins to turn on when \(V_{be}\)=0.6 and is fully saturated when \(V_{be}\)=0.8. So the active region is a voltage between those two points.
 

Audioguru

Joined Dec 20, 2007
11,248
Chips and Martin pretty much answered the question. The assumption is that the silicon transistor begins to turn on when \(V_{be}\)=0.6 and is fully saturated when \(V_{be}\)=0.8. So the active region is a voltage between those two points.
No.
That is an assumption that might be true for only a few transistors.
Vbe changes with the input current and with temperature.
The maximum Vbe of a 2N3055 power transistor with a 4A collector current is 1.5V, not 0.8V and is more when it is cold.
A transistor input is a current, not a voltage. Its input voltage requirement increases a little when its input current increases.
 

mlog

Joined Feb 11, 2012
276
No.
That is an assumption that might be true for only a few transistors.
Vbe changes with the input current and with temperature.
The maximum Vbe of a 2N3055 power transistor with a 4A collector current is 1.5V, not 0.8V and is more when it is cold.
A transistor input is a current, not a voltage. Its input voltage requirement increases a little when its input current increases.
Of course it is a rule of thumb and is typical of a small signal transistor. There are always exceptions to the "rule," which doesn't make what I said wrong. Thank you for your clarification.
 

Rick Martin

Joined Jun 14, 2009
31
For what Beijim is doing Vbe=0.6-0.8V is the value, there are ALWAYS variations however when learning and doing entry level small signal circuits Vbe=0.6-0.8V is accepted.
 
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