please check the attached file. i am stuck because i don't know how to come up with the third equation needed to find the three unkowns... Vce is .2, vbe(on) = .7, as the picture shows, i have written the KCL equations but don't know where to come up with the third equation? did i take the wrong approach to answer the question? thanks...
edit: please check the bottom drawing...sorry, i am a bit lazy to draw this on W-bench... i don't know where to come up with the third equation to find all the different values? i know for sure though from calculations i did that this is indeed in saturation mode... please look at the bottom transistor for clarity... thanks

 

Beta is used for a linear amplifier transistor that has plenty of collector to emitter voltage. The datasheet for almost every transistor shows its max saturation voltage when its base current is 1/10th its collector current, beta is not used.

You could guess that a "typical or better" transistor needs less base current but then some weaker transistors will not saturate.

 

guru, thanks for the reply. I think i understand what you say very well, but i am at a dead end though with the problem. it's not a concept i am missing- at least that's what i think...but i might have taken the wrong approach by writing the KCL equations??? maybe? could one please check if the approach i took to write the KCL equations is correct? if so, then what would be the third equations, because right now i have three unknowns with two KCLs...
thanks..
by the way, if anyone is interested the problem is listed on Niemen's Electronic Devices book... not the best book in terms of detailed understanding of why ICs work the way they do but excellent for problem solving skills... for the former i suggest Jager's book..jager lacks in problems or explanation of problems

 

I didn't check your work, but just looking quickly, I wonder if the following equation might help.

Ib+Ic=Ie

 

steveb, thanks... that was the third hint.... yes, that was the equation that i missed... concept wise, i don't know why i thought this, but i thought the current in the emitter WOULD NOT follow the same rule of thumb during saturation compared to forward-active or inverse mode.... so that means, the only thing that changes is hfe is not the same in saturation? and Vce becomes lower than vbe..

 

First we must check in which "state" the BJT is. Active or saturation.
So we assume
β=80; Vbe=0.7V; Et=6V; Ecc=5V

Ic=[ β* (Et-Vbe) ] / ( Rb + (β+1)*Re )= 4.8mA
Uce≈-20V
So transistor is in saturation.
Or differently way.
In this circuit the collector current cannot be greater than:
Icmax=Ecc / (Rc+Re)=1mA
So BJT is in saturation.
In saturation the Ic=β*Ib don't hold any more.

And now from KCL
Ib=(Et-(Vbe+Ve))/Rb
Ic=(Ecc-(Vce(sat)+Ve))/Rc
Ie=Ve/Re
Ib+Ic=Ie

\(Ve=(\frac{Et-Vbe}{Rb}+\frac{Ecc-Vce}{Rc})*(Rg||Rc||Re)=1.47V\)

Ie=Ve/Re=1.47mA
Ic=( Ecc-Vce(sat)+Ve )/Rc=832uA
Ib=(Et-Vbe+Ve)/Rb=638uA

Or
If you know Vce(sat) and Vbe then in the first group of equations
Ie=Ib+Ic
Rb*Ib+Vbe+Ie*Re=Et
Rc*Ic+Vce+Ie*Re=Ecc
you have only three unknown (Ie, Ic, Ib) and you can solve the system.
You can start substituting the first equation in the following two and obtaining.
Rb*Ib+Vbe+(Ib+Ic)*Re=Et
Rc*Ic+Vce+(Ib+Ic)*Re=Ecc