Ac load line do problem

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julian403

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For a transistor likes:



the DC load lines is:

\[Ic(sat) = \frac{Vcc}{Rc+Re}\]
and

\[Vce(sat) = Vcc\]

and the current is

\[Ic = - \frac {Vce}{Re+Rc} \]

And de AC load line:

\[ic(sat) =\frac{Vcc}{Rc+Re} + \frac{vce}{Rc//Rl}\]

So, that the same as

\[\frac{Vce}{x} =\frac{Vcc}{Rc+Re} + \frac{vce}{Rc//Rl}\]

What would x ?
 

Vce (sat) is the collector to emitter voltage when the transistor is turned on hard so it is saturated. It is not Vcc, instead it is very low, 0.1V or 0.2V.
The collector current when the transistor is saturated is not Vce/Re + Rc, instead it is Vcc/Re + Rc.

You are calculating DC currents. The AC currents will be different due to CE.
 
thank you to answer. I do not undertand, where would be Vceq and how it's the ic total current?

ic(sat) = Vcc/(Re+RC) + vce/(Rc+RL) that's the saturation current. right?
 

Vceq Icq can be anything you want. It is determined by the ratios of R1 and R2 and Rc and Re. Usually Vceq is designed for maximum symmetrical output swing.

Ic (sat) has nothing to do with Vce since Vce (sat) is very low 0.1V or 0.2V.
But AC sat current is different to DC sat current due to CE.
 
let me do the calculus.

\[VCE = Vceq + vce\]

where vce is AC potential difference between collector and emitter

\[iC = Icq + ic\]

where ic is the AC colector current. It's \[ic = - \frac{vce}{RL//Rc}\]

\[Icq= \frac{Vcc}{Rc+Re} - \frac{Vceq}{Rc+Re}\]

putting it together

\[ic = - \frac{vce}{RL//Rc}\]

\[iC - Icq = - \frac{VCE - Vceq}{RL//Rc}\]

so

\[iC = - \frac{VCE }{RL//Rc}+ \frac{ Vceq}{RL//Rc} + Icq\]

that's the ic current. it's right?
 

Why do you need so many calculations to draw a simple load line? School homework?
If the resistor values and power supply voltage were presented then you could easily see the quiescent collector voltage and quiescent collector current. Then since there is an emitter capacitor keeping the emitter voltage from changing it is simple to see the collector voltage rising to cutoff and dropping to saturation.
 
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