# Amplifier Circuit Question - Q of Ic

Discussion in 'Homework Help' started by howartthou, Jun 14, 2009.

1. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Hi All

I understand kirchoff's law, I undertsand basic transistor and amplifier theory, but try as I might I can't get anywhere with this circuit.

I try to analyse it and get nowhere.

The attached circuit has a question to determine the Q of Ic. I believe there is enough info to calculate Ic from Ie because Ie = Ic. So, if you can calculate Ie you will know Ic.

The problem I can't solve is how to calculate Ie when I have the value of E accross Re and the resistance of Re but I don't know the value of Rb2.

I know Ie = E / I based on ohms law, but which E? And how to I work out I for the Base-Emiter circuit?

If someone could explain how to work out the answer to the attached question I would really appreciate it!

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2. ### bertus Administrator

Apr 5, 2008
15,806
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Hello,

It is given that the power supply is 12 Volts.
It is given that the voltage across C-E is 6 Volts.
How many is left for the resistors in the collector and emitter?

Greetings,
Bertus

3. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Hi Bertus
If I answer your question with my current state of mind, which is almost fried right now, I would say 6v based on kirchoffs law.

Seems like I should know with this hint but I still get stuck.

My analysis so far:

Starting at ground 12v goes into RE. So, IRE = 12/330 = 0.036 A.

Ummm...stuck right here. How does this help me determine RB2 so I can try and determine IE?

I also assume I can ignore the 47ohm resistor because the circuit is open accross A and B.

To calculate IE I need to know the total resistance of the Base Emitter circuit don't I? RB2 is unknown, so how do I find that?

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Ie*330+6+Ic*1.5K=12

Ic=Ie

Thus,

Ic*330+6+Ic*1.5K=12

Ic(330+1500)=12-6

Thus,

Ic=6/1830=3.27mA

5. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Hi Mik
It seems obvious when you put it like that but still leaves me confused. You followed the circuit from ground to the power supply but I don't understand how you can ignore the RB1 and RB2, so:

1. Why did you ignore RB1 and RB2 aren't they in the circuit

2. Is your answer at the Q point?Is IE the same as IB which is the Q point

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
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If we apply KVL we get two independent mesh.
Input
Vcc=VRB1+VRB2=(I1+Ib)*RB1+ I1*RB2=Ie*Re+Vbe+Vcb+Ic*Rc

Just a little bit different
Vcc-I1*RB1-I2*RB2=0
I1=IB+I2
I2*RB2-Ube-Ie*Re=0
and thus,
Ib=[ (RB2*Vcc)-Ube*(RB1+RB2) ] / [ Re*(β+1)*(RB1+RB2)+(RB1*RB2) ]

and for the output loop
Vcc=VRc+Vce+VRe=Ic*Rc+Vce+Ie*Re.

And naturally Ib=Ic/β=Ie/(β+1)

Last edited: Jun 14, 2009
7. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Hi Jony
Seems like I asked for it didn't I?

I understand Input = Vcc=VRB1+VRB2.

I understand output = Vcc=VRc+Vce+VRe.

Thanks for clearing that up.

So I just need to use the output equation as mik3 described.

But:

1. Is Ic the Q point because its on the output loop so it is the half way point which is the Q point?

8. ### StayatHomeElectronics Well-Known Member

Sep 25, 2008
864
40
Analyzing at the Q point is just analyzing the amplifier as it is currently dc biased, no additional input in from Vi. You do not need to consider the Rb1 and Rb2 in the equation because of the conditions that they supply in the problem. Ic = Ie and Vce = 6 V. If you draw the power supply as a circuit element with the positive terminal going to the top (+12V) and the negative terminal going to the bottom (GND), then you may see the loop that is being analyzed more clearly.

You have a single current Ic going through both Rc and Re because of the condition Ic = Ie. Therefore Ic ( Rc + Re ) + Vce = 12 V as mik3 has shown.

9. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Thanks StayAtHomeElectronics

I did draw the power supply and its a bit clearer now. Jony has also helped with the same idea of adding in the power supply between +12v and ground. It does help me in the analysis.

Can I just ask about all those other ground symbols in the circuit:

1. Is the ground on RL connected to the other grounds like the ground on Vo and the grounf on Vi? Not sure how the grounds on Vi and Vo and RL connect to each other or not . Would appreciate and explanation here.

2. Shouldn't RL be connected directly to both terminals on Vo? Why is it shown detached out on its own like that?

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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1. All GND must be connected to the "minus" of the Vcc.
And if in diagram you see GND symbol, that mean that all gnd are connected to the same reference point.

2. Hmm, what is the difference between the connection.
If upper side of RL goes to Vo and the lower side goes to GND (the same point)

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11. ### mik3 Senior Member

Feb 4, 2008
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63
You assume that Ic=Ie and thus you can solve the KVL equation without the need to know the base voltage.

12. ### hgmjr Moderator

Jan 28, 2005
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The justification for the assumption that Ic = Ie stems from the transistor equation that Ie = Ic + Ib. Since Ib = Ic/beta, then with beta values in the range of 100 or more you can safely disregard Ib in the transistor equation.

hgmjr

13. ### howartthou Thread Starter Active Member

Apr 18, 2009
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Excellent Jony, thanks!

I appreciate the diagram, it has really helped me alot. I also found some reading material on Superposition, loop analysis and nodal analysis.

And, if I read your version of the diagram (which is the clearest version) I think starting at ground (0v) the current goes through RL and then C2 and then Rc and then the +12v EMF.

Is this the right way to read it?

I think you would recommend I start at EMF and work my way to RL and then ground, right?

14. ### howartthou Thread Starter Active Member

Apr 18, 2009
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hgmgr

Thanks, very insightful. Just what my book would say....

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, I prefer this version