The problem is that you won't quit while you're ahead. You insist on then imposing (for part (a)) a condition that does not exist. The current in the collector is NOT 7.5 mA. Is it also not zero. It is something in between. You need the beta of the transistor in order to determine the collector current in the transistor and then you use the load line to determine the corresponding Vce, which will be somewhere between o V and 5 V.

so i should go a step backward and assume beta for part a to get the right value of ic and it must be between zero and 5v which is Vthev
assuming β = 50 then ic = 50 * 43uA = 2.15 mA
Vce = Vthev - (ic*Req)
Vce = 5 - ( 2.15mA * 1000 )
Vce = 2.85v
is this correct ?
if it is right so irl =[ 2.85 - ( -5) ] / 2k = 3.92mA

The current in the collector is NOT 7.5 mA.

but i remember u told me that ic sat = 7.5mA
every time i think i got the solution u tell me that there is a mistake so i go back to where i started i feel hopless

 

so i should go a step backward and assume beta for part a to get the right value of ic and it must be between zero and 5v which is Vthev
assuming β = 50 then ic = 50 * 43uA = 2.15 mA
Vce = Vthev - (ic*Req)
Vce = 5 - ( 2.15mA * 1000 )
Vce = 2.85v
is this correct ?
if it is right so irl =[ 2.85 - ( -5) ] / 2k = 3.92mA

That looks correct IF beta is 50. We don't know what it is. So that brings us back to those three options I've mentioned a couple of times. You've just chosen the first one. That may be good enough (I don't know what your instructor expects you to do, so it may not be).

but i remember u told me that ic sat = 7.5mA
every time i think i got the solution u tell me that there is a mistake so i go back to where i started i feel hopless

Icsat IS 7.5 mA. But there is absolutely no reason to believe that the circuit in part (a) is in saturation and, in fact, there is reason to believe it is not. The circuit in part (b) IS in saturation because they are asking you to find the value of Vbb that makes this happen.

This is what I mean by field dependence. I suspect that if you had never seen part (b) of this question, you wouldn't be struggling as much. But you did. So now the information related to part (b) is part of the context of the problem for you and you are having a hard time separating them back apart.

 

no i'm not like that believe me i tried a lot and i'm still trying and i never wanted to ask for help but really i feel i'm so fool now
i'm always the best in analysing circuits but i don't know why i'm completely lost to this one i can't find it even though i feel it is too close to me but i can't see it

why did u put vce0 in the equation ?? and what difference is it gonna make ??
please don't think i'm stupid but this circuit isn't in my level
i just like to learn faster

Vce0 is simply the value of Vce when Ic is zero. Nothing more than that. It's just a name for the y-intercept of that equation.

 

That looks correct IF beta is 50. We don't know what it is. So that brings us back to those three options I've mentioned a couple of times.

i know but i wanted to ask first if i'm going in the right way

So that brings us back to those three options I've mentioned a couple of times

u mentioned three choices i used the first and the next was to leave the answer parameterized in terms of β which i don't prefer and the third was to Provide plots of the answer over a range of β
i can assume many values foee beta
assume beta = 10 then ic = 10 * 43uA = 430uA
Vce = Vthev - ( ic * req )
vce = 5v - 0.286 = 4.714v
assume β = 20 then ic = 20 * 43uA = 860uA
vce = 5 - ( 860uA * 667 ohms ) = 4.427v
assume β = 30 then ic = 1.29 mA
vce = 4.14v
assume β = 40 then ic = 1.72mA
vce = 3.86v
assume β = 50 then ic = 2.15mA and vce = 3.57v [ and here i noticed that i made a mistake back there with req and said req = 1000 but it is 667 ohms ]
assume β = 60 then ic = 2.58mA and vce = 3.28v
assume β = 70 then ic = 3mA and vce = 3v
assume β = 80 then ic = 3.5mA vce = 2.7v
assume β = 90 then ic = 3.8mA vce = 3.5 v
assume β = 100 then ic = 4.3mA vce = 1.14v
assume β = 110 then ic = 4.73mA vce = 1.84v
assumr β = 175 then ic = 7.5 mA and vce = 0v
but does this mean that β will remain as a parameter no matter how i try ?? and so this is how am i gonna end this circuit ? we didn't deal with circuits like these before
to leave β like this is something new to me
there is no way i can find a specific value for it

Icsat IS 7.5 mA. But there is absolutely no reason to believe that the circuit but this means that i can't in part (a) is in saturation and, in fact, there is reason to believe it is not. The circuit in part (b) IS in saturation because they are asking you to find the value of Vbb that makes this happen.

now i understand what was my fault about ic i was really so close of the right solution but only one mistake made me spend a week trying
my fault was to say Vce = Vthev - (ic SAT * req )
since i posted my last reply i thought there is some thing wrong with this equation and i found it immediately [ but had some troubles with the internet ] how am i gonna get vce in active while using ic sat ?
this equation is the first thing i did in this circuit even before ib when i was thinking that it's too easy and didn't take it seriously

 

i emailed the person who posted the circuit a week ago to ask him about it but i didn't get my answer until now and i fell ohhhhh because the person said that β was posted two days after that because this circuit is a series of similar circuits he kept posting for a week and for all the circuits β is 100
so all those days are for nothing
any way thank to u i learnt a lot of new things and the solution now is :
for part a
ib - 43uA
ic= 4.3 mA
vce = Vthevv - ( Req * ic )
vce = 5v - 2.86v
vce = 2.13v
irl = [ 2.13v - (-5v) ] / 2k = 3.5mA
for part b
ic sat = 7.5mA
vce sat = 0v
for ic = 7.5 mA and β = 100 ib must be
ib = ic / β = 7.5mA / 100 = 75uA
vbb = (rb * ib ) + 0.7v
vbb = ( 100k * 75uA )+ 0.7v
vbb = 8.2v

 

i emailed the person who posted the circuit a week ago to ask him about it but i didn't get my answer until now and i fell ohhhhh because the person said that β was posted two days after that because this circuit is a series of similar circuits he kept posting for a week and for all the circuits β is 100
so all those days are for nothing

From Post #4: "Look again to see if they don't indicate a value of beta to use. Perhaps they tell you somewhere what value to use as a default."

But if you learned some things, it wasn't for nothing.

any way thank to u i learnt a lot of new things and the solution now is :
for part a
ib - 43uA
ic= 4.3 mA
vce = Vthevv - ( Req * ic )
vce = 5v - 2.86v
vce = 2.13v
irl = [ 2.13v - (-5v) ] / 2k = 3.5mA
for part b
ic sat = 7.5mA
vce sat = 0v
for ic = 7.5 mA and β = 100 ib must be
ib = ic / β = 7.5mA / 100 = 75uA
vbb = (rb * ib ) + 0.7v
vce = ( 100k * 75uA )+ 0.7v
vce = 8.2v

Do you really mean that Vce is 8.2 V in that last line?

 

Do you really mean that Vce is 8.2 V in that last line?

no sorry not vce but vbb
changed it

 

From Post #4: "Look again to see if they don't indicate a value of beta to use. Perhaps they tell you somewhere what value to use as a default."

looked for it a lot but it wasn't written on that page

 

looked for it a lot but it wasn't written on that page

In the future, be a bit quicker to request clarification. That goes not only for homework, but in the "real world", too. You bosses and your customers aren't always going to give you all the information you need and it usually will not be intentional. They may not realize you need it or they may not realize they didn't actually give it to you. The worst that can happen (usually) is that they tell you that you need to deal with it on your own.

 

In the future, be a bit quicker to request clarification. That goes not only for homework, but in the "real world", too. You bosses and your customers aren't always going to give you all the information you need and it usually will not be intentional. They may not realize you need it or they may not realize they didn't actually give it to you. The worst that can happen (usually) is that they tell you that you need to deal with it on your own.

i will thanks a lot for ur precious advises and for helping me all these days