# BJT amplifier

Discussion in 'Homework Help' started by yoamocuy, Oct 25, 2010.

1. ### yoamocuy Thread Starter Active Member

Oct 7, 2009
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Based on the circuit design shown, I am supposed to find the following:

2) What happens to Av, Gv, Ai, and Gi when Rc changes.

3)max value of Rc and why?

4) min value of Rc and why?

5) What happens Av, Gv, Ai, Gi, hie, and ro when Rb changes?

6) max value of Rb and why?

7) min value of Rb and why?

I've managed to find everything but the max and min values of Rc and Rb. I found that Av max is equal to -Vcc/VT so I did find a max value of Rc and a min value of Rb that sets Av at its max value but I'm not even positive if that is correct and I have no idea about the min value of Rc and max value of Rb. Is there a minimum value of Ai or Av? Or am I completely off in my pursuit of this question?

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2. ### PRS Well-Known Member

Aug 24, 2008
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Last edited: Oct 25, 2010
3. ### yoamocuy Thread Starter Active Member

Oct 7, 2009
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Ok, I'll show all the work I've done so far and state specifically what problems I am occurring so you can see that I am not simply trying to get answers.

1) I attached a photo of my load lines. I'll go ahead and explain how I found them below.

I made a voltage loop around the base and came up with the equation Vb=Rb*ib+hie*ib+Re*ie+Vbe. Since ie=ib*(B+1), hie=25m*100/icq and ib=ic/100, I rewrote the equation as follows. icq/100=(Vb-Vbe)/(Rb+25m*100/icq+100*101). Plugging in 5 for Vb, 0.7 for Vbe, and 100 for B I came up with icq=5mA.

To find the D.C. Load Line, I used the equation Vcc=ic*(Re+Rc)+Vce. If Vce=0V, then ic is equal to 20mA because Re is 100 and Rc is 900. If ic=0mA, then Vce is 20V. My Vce at operating point was then found by plugging in 5mA for ic since that was what my icq was found to be, and then solving for Vce.

To find the A.C. Load Line, I first used the equation IcqAC=icq+Vce/RAC, where RAC is 900 because that is the resistance at the output of the transistor in AC. This equation gave me an icqAC of 21.67mA. I then used the equation VceAC=Vce+icq*RAC. This equation gave me a VceAC of 19.5 V.

2)Using the equations listed at the top of the problem, it is apparent that as Rc gets larger, Av and Gv get larger while Ai and Gi get smaller.

3)For the max value of Rc, I first found Avmax to be -800 based on the equation Avmax=-Vcc/VT. I then set -ro||Rc(B)/hie=-800 and solved for Rc. Using this equation I got a value of Rc to be 5k. Having an Rc of this value would change the D.C. Load Line in such a way that the circuit would then be in saturation so the icq would change so this can't be right. I can find the value of Rc that will get the circuit close to saturation but really the Rc could be higher than that value so I'm not sure what about that. I could also find the largest value of Rc that would allow for the circuit to have a voltage swing that doesn't clip, but I could still increase the Rc it would just cause the output to clip.

4)For the min value of Rc, I'm not really sure what to do or really what to think. I have the same dilemma as I do with the max, except I don't have any equation for Aimax as I do for Avmax.

Honestly, if I can figure out what the max and min values of Rc are and why, then I can figure out Rb max and min without a problem. The circuit analysis for this one is extremely simple, I'm just struggling with that portion. Any advice would be appreciated.

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4. ### hobbyist Distinguished Member

Aug 10, 2008
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This person, has gone through a lot of calculations, and it is VERY apparent this person is NOT trying to just get answers to a homework problem, as people in the past has done.

Please can someone, who is knowledgeable in this give this person some help.

Or if you are working on the problem, please at least acknowledge, so this person doesn't have a discouraging experiance with this forum.

Thankyou.

5. ### R!f@@ AAC Fanatic!

Apr 2, 2009
8,523
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I am a bit rusty on the theory part. I did a lot of problems on amp and load line analysis during my teaching days, but now somethings are forgotten and it just doesn't pop up in my mind now like they use to.

I guess OP has to wait till someone fresh finds this thread.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,938
1,089
For this simple amplifier we can determine.
Vbb = Ib* Rb + Vbe + Ie * Re

Ie = Ib *
β + Ib = (β+1) * Ib

Ibq = ( Vbb - Vbe) / ( Rb + (
β+1)*Re ) = (5V - 0.7V) / 85.6K = 50.23μA

Icq =
β*Ibq ≈ 5mA

Vceq = Vcc - Ic*Rc - Ie*Re
Vcc - Ic*(Rc+Re) = 15V

Not very good choice becaues of pure output voltage swing.
For a given Rc and Re with Ce capacitor.
To get maximum voltage swing for a given Vcc, Icq must be equal to:

$Ic_{opt}=\frac{Vcc-Vce(sat)}{2Rc+Re}= 10.4mA$

The voltage gain is equal for rsis = 0Ω

Av = gm* Rc = Ic/26mV * Rc = 173 V/V

Current gain
Ai = β * Rb/ ( Rb + Hie) ≈ 99

Hie = (β+1) * 1/gm = 101 * 26mV/5mA = 525.2Ω

If we change Rc to 0Ω then Voltage gain --> 0 V/V
But for Rc = ∞ Voltage gain is equal to VA / Vt = 100V/26mV = 3.8K [V/V] VA - Early voltage
Vt - thermal voltage

And Rc also have some influence on current gain Ai.
If we take into account ro then current gain is equal

Gi = Rb/ ( Rb + Hie) * β * ro/( ro + Rc)

For a given bias point (Icq = 5mA) we can increase the RC until BJT will star to saturation.
If we assume the saturation begins at Vc= Vb then
Rc_max = ( Vcc - Vb) /Icq = 3.75K
Ω
But for this Rc output voltage will be very distorted.

We can use Rc=0Ω replace Rc with a short circuit, but then output voltage is equal Vcc so voltage gain is equal 0V/V.
So it is good to have some voltage gain in voltage amplifier don't you think.
Decreasing Rc decreases voltage gain but also decreases Rout.

Change in Rb has a large influence on current gain

Rb/ ( Rb + Hie) * β

and off course on voltage gain to (for rsis >> 0 ) voltage gain is equal

Gv = (Rb||Hie) / ( rsis + Rb||Hie ) * gm*Rc||ro

Increasing value of a Rb resistor we shifts the operating point toward cut-off.

And by decreases Rb value we move operating point toward saturation.
Ic_max = ( Vcc - Vce(sat) )/ ( Rc+ Re) = 19.8mA

Vb = 19.8mA*100Ω = 1.98V
Ib = 19.8mV/β = 198μA

Rb_min = ( 5V - 1.98V)/ 198uA = 15.2KΩ

7. ### yoamocuy Thread Starter Active Member

Oct 7, 2009
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0
Ok, I thought that could be all the min's and max's depended upon. Thanks for the clarification.

8. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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I'm late to the thread. If you have another occasion to start a new thread/new question, please do include how far you've gotten, showing the work like you did in your second post above.

It is nothing against you, but too many people stop by to just get answers. We are more than happy to help people learn, and simply giving answers does more harm than good, as knowing the answer to one question doesn't help one bit if you don't know why it is the answer.

Sorry it has come to this, I take breaks now and again, half wishing the Internet was around when I was in college for EE, half wishing college profs would change their questions more often so they couldn't be googled.

Check the link in my sig for making your work look GOOD. Clearly posting your formulas sometimes saves a good deal of confusion.

Lastly, welcome!