# Common-emitter amplifier question

Discussion in 'Homework Help' started by beijim, May 11, 2012.

1. ### beijim Thread Starter New Member

May 11, 2012
3
0
First, sorry for my poor English.

The question is shown below.

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 !

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 ...

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

Can anyone help me?

Finally, thanks for you help.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
1,088
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.

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3. ### MrChips Moderator

Oct 2, 2009
12,229
3,281
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: May 12, 2012
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4. ### beijim Thread Starter New Member

May 11, 2012
3
0

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

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: May 12, 2012
5. ### Rick Martin Active Member

Jun 14, 2009
31
2
It all looks correct now although BETA is extremely high.

Also Vce = 6.0V is given, why are you questioning this?

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6. ### Audioguru New Member

Dec 20, 2007
9,411
895
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.

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7. ### beijim Thread Starter New Member

May 11, 2012
3
0
Oh...I would like to know why Vbe=0.6V but typed Vce=6V...sorry.

8. ### MrChips Moderator

Oct 2, 2009
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9. ### Rick Martin Active Member

Jun 14, 2009
31
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0.6V is the voltage lost across a silicon P-N junction. Vbe = 0.6 - 0.7 is a standard assumption made.

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10. ### mlog Member

Feb 11, 2012
276
36
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.

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11. ### Audioguru New Member

Dec 20, 2007
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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.

12. ### mlog Member

Feb 11, 2012
276
36
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.

13. ### Rick Martin Active Member

Jun 14, 2009
31
2
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.