# Find the voltage gain

Discussion in 'Homework Help' started by ADCapacitor, Mar 5, 2016.

Feb 3, 2016
42
0
β=210, Vt=26mV, Vbe(on)=0.7V
How to find Vs,
then find the Voltage gain Vo/Vs ?
I try to draw the small signal model to solve the problem

Vth=15*R2/[R1+R2]
Vth=1.62V

Rth=R1//R2
=5.5kΩ

-Vth+Rth*Ib+Ie*Re+Vbe=0
Vth=Rth*Ib*(1+β)*Re
Ib=(Vth-Vbe)/Rth+Re(1+β)
=8.2μA

Since
Re is short in Ac

Rπ=Vt/Ib
=3.18k ohm

Vo=-β*ib*Rc
How to find Vs,
then I can get Vo/Vs ?

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Why Vo is at emitter ??

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,519
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Current amplifier?

The better question is how resistor has a value of 10 uF. On the other hand that same resistor is behind dc blocking capacitor so it likely does not matter.

Ok. Question to OP. What text are you using for learning these stuff?
My textbook has some good examples on these things and explanations for the approximations that can be used to shortcut thought the complicated math.

4. ### anhnha Active Member

Apr 19, 2012
776
48
With current amplifier or buffer I think it is better not to include the collector resistor.

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
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You are most likely right. I assume that we are dealing with textbook exercise, therefore the actual exact function of the circuit does not matter.

Feb 3, 2016
42
0
Thank you
I make some mistakes when redrawing the circuit diagram
The resistor with wrong label is the internal resistor of the signal generator.And the value is not given.
I think this one is better

Feb 3, 2016
42
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I am using the lecture notes for learning these stuff.
My teacher give us this question for practice
Which textbook are you using?It sounds good

8. ### WBahn Moderator

Mar 31, 2012
18,096
4,920
I'm not at all sure that this is supposed to be a current amplifier. Even if it is, the collector resistor could be there to either put a limit on the current output or to reduce the collector-emitter voltage to something that compensates for the Early effect.

9. ### shteii01 AAC Fanatic!

Feb 19, 2010
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515
My textbook was Electronic Circuit Analysis and Design, Second Edition, Donald A. Neamen.

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Feb 3, 2016
42
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The output should be at collector

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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So you have a traditional Common Emitter amplifier. And your problem is Vs ?? So first try to draw the "correct" small-signal model.
An then you will see that if internal resistor of the signal generator Rs = 0Ω Vin = Vs = voltage across rpi.

Feb 3, 2016
42
0

Is that correct?
Vs=Vin
Vin=Ib*R1//R2 ?
Vo=-βIb*Rc

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Are you sure that Ib current is flowing through R1||R2 ?? If so, which current is flowing through rpi then?

Feb 3, 2016
42
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May be this one
Vs = ib*Rpi ?

16. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yep, this time you are on the right track.

Feb 3, 2016
42
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thank you
I have some further questions
I found that some examples use this "RTH=0.1(1+β)Re" to find the Rth
And it sometimes use Rth=R1//R2
But why?

18. ### Jony130 AAC Fanatic!

Feb 17, 2009
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When we want to find the DC operation point (quiescent current) in order to simplify the calculation we use thevenin's theorem and replace Vcc and voltage divider at the base by his thevenin equivalent. And in this case Rth = R1/R2 and Vth = R2/(R1+R2)
Can you show us the example? Because (1+β)Re is a AC resistance seen from the base terminal when we looking into emitter and if we ignore r'e.

Feb 3, 2016
42
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Design a bias-stable circuit . Let β = 100 and VBE(on) = 0.7 Vm , Ic=0.7mA, Vce=4V

this example use Rth=0.1(1+β)Re to find Rth
Is "Bias-stable" imply that we can use Rth=0.1(1+β)Re to find the Rth?

20. ### Jony130 AAC Fanatic!

Feb 17, 2009
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OK I see. This formula "Rth=0.1(1+β)Re" is a rule of thumb used sometimes by a designer. And this rule means that we must choose R1||R2 < 0.1(1+β)Re for "bias-stable". This rule ensure that our voltage divider voltage is "stiff". Because base current is a load for our voltage divider. And to ensure that our divider voltage do not change much after we loading it we use this "rule of thumb" to select R1 and R2. The slightly different rule of thumb is to choose voltage divider current Id = Vcc/(R1+R2) at least 10 times the base current. So we have:
R1 = (Vcc - Vb)/(11*Ib)
R2 = Vb/(10*Ib)