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#1
03-19-2010, 04:37 AM
 cheypr Junior Member Join Date: Oct 2009 Posts: 34
How to find Q point load line

Hi again, I'm trying to graph the Q point and load lines for this class A amp.
From what know Vce(max)=Vcc and Ic(max)=Vcc/Rc.

Do I got it right? How do I get Vce max and Ic max?

#2
03-19-2010, 06:23 PM
 The Electrician Senior Member Join Date: Oct 2007 Posts: 1,748

Everything looks ok; I didn't check all the numerical results. You've made some approximations, such as in the calculation for VB where you didn't take into account the effect of the base current on the voltage divider, etc., but approximations are only approximate, right?

The one thing I see that doesn't look quite right is at the very end where you calculate Ic(max).

If you want Ic(max) when the transistor is held full on with DC drive, then Ic(max) would be something like Vcc/(100+8.2+36).

If you want Ic(max) given the 500 mV P-P input signal, that's a different calculation!

But, the problem didn't ask for Ic(max), so you're ok.
#3
03-19-2010, 09:21 PM
 PRS Senior Member Join Date: Aug 2008 Location: Yakima Washington Posts: 989

Notice the DC load line is asked for so I simplified the schematic accordingly. Follow the equations step by step and you'll see the reasoning. This is approximate. Your teacher probably wants you to use beta and/or alpha and a more exact solution for Ie. Remember the loop equation for that? More on this point if you want.
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Paul
Yakima, Washington
#4
03-20-2010, 01:52 AM
 The Electrician Senior Member Join Date: Oct 2007 Posts: 1,748

Reading PRS's post prompted me to go back and have another look at your work. I obviously read through it too fast!

In the second column, you have Vc =Vcc - Ic*Rc - IE*RE

You should have stopped there, and let the equation be:

Vce =Vcc - Ic*Rc - IE*RE

because when you went further, with Vce = Vc - VE, you ended up subtracting VE twice from Vc.

You would have ended up with Vce = 5.21, approximately correct.
#5
03-20-2010, 06:34 PM
 PRS Senior Member Join Date: Aug 2008 Location: Yakima Washington Posts: 989

My evaluation was a little crude, given the relatively high amount of current going into the base. You need to thevinize the input as an equivalent resistance and voltage, then write a loop that includes that, Vbe and Ie, solving it for Ie. This will give you a more exact Ve and Vc. Then use these to determine Vce = Vc - Ve. Get Ic from Beta and Ic(sat)is solved as above. Ic(sat), by the way is the total amount of current available from the source given Vcc and the resistances of the circuit. Draw the load line the same as I drew it. This should put your teacher in tall cotton when he evaluates your homework!
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Paul
Yakima, Washington
#6
03-21-2010, 04:25 AM
 cheypr Junior Member Join Date: Oct 2009 Posts: 34

Ok, thanks to both of you guys!!!!!

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