Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Ac load line do problem

Status
Not open for further replies.

julian403

Full Member level 5
Full Member level 5
Joined
Feb 28, 2014
Messages
254
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Location
Argentina
Activity points
2,105
For a transistor likes:

asdf.jpg

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.
 
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top