Ic = I1 + Irl

Is this correct?

i was chocked when i read this but i went back to the figure and saw it
that's right it's in the opposite direction
i didn't think that it can go in this direction because in this case it'll be moving from a low potential point to a high potential one but then i realised that the positive of the battery is grounded
when i was trying to solve i was focusing on vce and next i thought about ic but i used my memory drawing the circuit on papers many times thinking that it wasn't grounded
well now it's clear that i1 = ic + irl
then ic = i1 - irl and here is the problem again where i1 and irl both = 5 mA !!!
so turned around and got ic = zero again eventhough i'm sure it's wrong
don't tell me that ic = i1 then where did irl come from ??
now where is the problem

Then why did you go right on ignoring them in this very post?

For example:

Ic = 0.005 + 0.005 = 0.010 A or 10 mA

0.005 is just a number. It is NOT a current. This should be

Ic = 0.005 A + 0.005 A = 0.010 A or 10 mA

u r right fell again in the same fault but actually not because i forgot this time i thought that since the two numbers are with the same units so no need to mention it but u r right about that u convinced me i always think it's faster to ignore it but i shouldn't do that again

 

Hi,

Yes in your problems that would mean that the input voltage either has to be 0.7 more or 0.7 less depending on connections. So after calculating the base current you would have to figure in the 0.7v drop along with the base resistor.

yeah that's what i did

 

With some experience you'll develop a feel for when you can ignore these (assume they are both zero), use these fixed values, or have to use something better such as Vbe changing by about 60 mV for each decade change in collector current.

didn't know that i thought it's constant

In most circuit design and analysis, particularly at your stage of learning, a good general rule is to make a first pass assuming Vbe and Vcesat are both zero. That usually allows you to design or analyze the circuit considerably quicker and get an answer that is close. That's particularly valuable when you are designing the circuit because you will probably be going through several iterations

we did it for a while but then we knew that it's not actually zero and started analysing circuit depending on that

Once you are ready for the real thing, you can use a better model (such as fixed values, which are often "good enough") and if the results vary considerably from your zero volt model, then there is a very good chance that either you have made a mistake somewhere, or that your circuit is so sensitive to these parameters that the fixed-value model is probably NOT good enough. In either case, you need to look more closely.

well i'm told that these values are not fixed for all transistors but i didn't know that they vary due to the current
any way what's the best way to deal with these parameters ? should i keep assuming that vbe = 0.7v and vce = 0.2 v or what ??

 

i was chocked when i read this but i went back to the figure and saw it
that's right it's in the opposite direction
i didn't think that it can go in this direction because in this case it'll be moving from a low potential point to a high potential one but then i realised that the positive of the battery is grounded
when i was trying to solve i was focusing on vce and next i thought about ic but i used my memory drawing the circuit on papers many times thinking that it wasn't grounded
well now it's clear that i1 = ic + irl
then ic = i1 - irl and here is the problem again where i1 and irl both = 5 mA !!!
so turned around and got ic = zero again eventhough i'm sure it's wrong
don't tell me that ic = i1 then where did irl come from ??
now where is the problem

You are making progress. The problem here is one that I have been trying to point out for some time. In this analysis you assumed that Vce was 5 V. Well, IF Vce is 5 V, the Ic is zero. The problem is that there is no reason to claim that Vce is 5 V and, in fact, compelling reason to claim that it is NOT. It is an unknown. All you have done up to this point is establish the load line, which is the relationship between Vce and Ic -- which is why I keep trying (and thus far unsuccessfully) to get you to come up with the equation that related Vce and Ic. You know that the Thevenin equivalent is a linear circuit, so the voltage and current change linearly. You know two points on the curve, namely (Vce, Ic) = (5 V, 0 mA) and (0 V, 7.5 mA). All you know is that the actual values for Vce and Ic are somewhere between these points. In order to determine where, you need to know the beta of the transistor. Which brings me back to another thing that I tried to tell you before but that perhaps you have made enough progress to be more receptive to at this point. You basically have three choices.

1) Assume a value of β.
2) Leave your answer parameterized in terms of β.
3) Provide plots of your answer over a range of β.

 

didn't know that i thought it's constant

we did it for a while but then we knew that it's not actually zero and started analysing circuit depending on that

well i'm told that these values are not fixed for all transistors but i didn't know that they vary due to the current
any way what's the best way to deal with these parameters ? should i keep assuming that vbe = 0.7v and vce = 0.2 v or what ??

You have a few more stages of learning to go and at each one you will learn that life isn't as good as you thought it was. Soon you'll discover that not only do some of these parameters, particularly beta, vary considerably from one type of transistor to another, but they vary between different transistors of the same type, even from the same die. They also vary with current and voltage. They also vary with temperature. They also vary with age. As a result, designs (like the ones in these problems) that rely on a known and constant value of beta are bad designs. You'll learn techniques to design circuits so that the behavior of the circuit is nearly unchanged even as the value of beta changes by over an order of magnitude. You may even learn how to take advantage in how some of these parameters change in order to make circuits perform better or to do things that they couldn't do if these parameters truly were fixed.

But, for now, let things develop at the pace that your courses are developing them at. The intent is to baby step you down a path of concepts so that you can learn and become proficient at them one or two at a time without getting overwhelmed.

 

You are making progress. The problem here is one that I have been trying to point out for some time. In this analysis you assumed that Vce was 5 V

progress ? that made laugh and cry what progress r u talking about if u r telling me that vce isn't 5v ??

Well, IF Vce is 5 V, the Ic is zero. The problem is that there is no reason to claim that Vce is 5 V and, in fact, compelling reason to claim that it is NOT. It is an unknown

so what is it 5v or not ??? now i'm not sure of any result i got please tell me what's wrong of my answers and what's right ?? ic at saturation = 7.5mA wrong or right ??
vce = 5v wrong or right ??
ic in active i'm still not sure about
ib = 43uA wrong or right ??
irl = 5mA wrong or right ??

All you have done up to this point is establish the load line, which is the relationship between Vce and Ic -- which is why I keep trying (and thus far unsuccessfully) to get you to come up with the equation that related Vce and Ic

so all i did was nothing ? every thing is wrong ??
i don't know if u mean the relationship like this but i just thought about it : ic is i input for the lod=ad line and vce is v ino=put for the load line so it's :
if vce = 5v then ic = irl = 5mA and i'm not sure that it can be right

In order to determine where, you need to know the beta of the transistor. Which brings me back to another thing that I tried to tell you before but that perhaps you have made enough progress to be more receptive to at this point. You basically have three choices.

1) Assume a value of β.
2) Leave your answer parameterized in terms of β.
3) Provide plots of your answer over a range of β.

but i did it i assumed a value for beta here :

at saturation : vce = 0v ic = 7.5 mA beta = ? ib = ic / beta
assume thet beta = 50 then :
vbb = ib * rb - 0.7v ib = ic / beta
therefore : vbb =[(ic * rb ) / beta ] = 750 / 50= 15
to make that this value of beta is good :
ib = (15-0.7) / /100000 = 0.000143
ic = ib * beta = 0.000143 * 50 = 7.15 mA a close value but maybe i need to try another one
i tried beta = 15 and got vbb = 50 v and a very close value to ic 7.39 mA
did i do right this time ??

part (b) :
at saturation : vce = 0v ic = 7.5 mA beta = ? ib = ic / beta
assume thet beta = 50 then :
vbb = ib * rb - 0.7v ib = ic / beta
therefore : vbb =[(ic * rb ) / beta ] = 750 / 50= 15
to make that this value of beta is good :
ib = (15-0.7) / /100000 = 0.000143
ic = ib * beta = 0.000143 * 50 = 7.15 mA a close value but maybe i need to try another one
i tried beta = 15 and got vbb = 50 v and a very close value to ic 7.39 mA
did i do right this time ??

was it wrong ?? is ic = 7.5mA wrong ??

 

You have a few more stages of learning to go and at each one you will learn that life isn't as good as you thought it was. Soon you'll discover that not only do some of these parameters, particularly beta, vary considerably from one type of transistor to another, but they vary between different transistors of the same type, even from the same die. They also vary with current and voltage. They also vary with temperature. They also vary with age. As a result, designs (like the ones in these problems) that rely on a known and constant value of beta are bad designs. You'll learn techniques to design circuits so that the behavior of the circuit is nearly unchanged even as the value of beta changes by over an order of magnitude. You may even learn how to take advantage in how some of these parameters change in order to make circuits perform better or to do things that they couldn't do if these parameters truly were fixed.

But, for now, let things develop at the pace that your courses are developing them at. The intent is to baby step you down a path of concepts so that you can learn and become proficient at them one or two at a time without getting overwhelmed.

i understand but when u said

You'll learn techniques to design circuits so that the behavior of the circuit is nearly unchanged even as the value of beta changes by over an order of magnitude

did u mean that these circuits i'm dealing with now or this level of circuit won't be used when i become more proficient ??

 

i understand but when u said

did u mean that these circuits i'm dealing with now or this level of circuit won't be used when i become more proficient ??

That's correct (by an large). These circuits are useful for learning the fundamentals and how to work with them in a greatly simplified world.

Have you had any course in physics (mechanics) If so, then you spent a lot of time working with frictionless surfaces and pulleys, cannonballs that experience no drag, and perfectly inelastic collisions and, at the time, you were more than happy just going along unaware that friction was going to be thrown into the mix in a few weeks and life would get quite a bit more complicated. Now consider that in the real world, frictionless pulleys don't exist and drag can't be ignored. So why did you spend so much time ignoring it? Because it let you focus on the fundamentals because having to contend with more realistic situations.

 

progress ? that made laugh and cry what progress r u talking about if u r telling me that vce isn't 5v ??

so what is it 5v or not ??? now i'm not sure of any result i got please tell me what's wrong of my answers and what's right ?? ic at saturation = 7.5mA wrong or right ??
vce = 5v wrong or right ??
ic in active i'm still not sure about
ib = 43uA wrong or right ??
irl = 5mA wrong or right ??


so all i did was nothing ? every thing is wrong ??
i don't know if u mean the relationship like this but i just thought about it : ic is i input for the lod=ad line and vce is v ino=put for the load line so it's :
if vce = 5v then ic = irl = 5mA and i'm not sure that it can be right

but i did it i assumed a value for beta here :



was it wrong ?? is ic = 7.5mA wrong ??

Not all of it was wrong. Much of it is now right.

As for where you assumed a value for beta, that was in part (b). We are still discussing part (a). That's why it is such a bad idea to have more than one problem in a thread -- the conversation invariably becomes to convoluted that it is nearly impossible to follow.

You ib = 43 μA is correct.

Let's try this.

Assume that I tell you that, in part (a), the beta of the transistor is 100. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

Now let's assume that I replace the transistor with one whose beta is 20. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

Now let's assume that I replace the transistor with one whose beta is 150. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

Now let's say that I replace the transistor with one whose beta is β. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

 

Assume that I tell you that, in part (a), the beta of the transistor is 100. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

well if i assumed beta from this stage it'll be too easy ic will be equal to beta * ib then ic = 100 * 43 uA = = 4.3 mA
and vce = 10v - 1k * 4.3 mA = 10v - 4.3v = 5.7v and this is without thinking about the load line which is in parallel and it might effect on this value

Now let's assume that I replace the transistor with one whose beta is 20. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

ic = 20 * 43uA = 860 uA
vce = 10v - 1kohms * 860 uA = 9.14v

Now let's assume that I replace the transistor with one whose beta is 150. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

ic = 150 * 43uA = 6.45mA
vce = 10v - 1k ohms * 6.45mA = 3.55v

Now let's say that I replace the transistor with one whose beta is β. Can you now work it and come up with a value for Vce and Ic? What is the value of Irl?

ic = β * 43uA
vce = 10v -1kohms * (β * 43uA ) so i left β as a parameter

 

well if i assumed beta from this stage it'll be too easy ic will be equal to beta * ib then ic = 100 * 43 uA = = 4.3 mA
and vce = 10v - 1k * 4.3 mA = 10v - 4.3v = 5.7v and this is without thinking about the load line which is in parallel and it might effect on this value

You started off on the right track, but just because Ic is 4.3 mA does NOT mean that the current in the 1 kΩ resistor is 4.3 mA.

The whole point of developing the Thevenin equivalent as seen by the collector-emitter of the transistor is so that you could easily address this part of the problem. That is why I keep asking you to come with with the equation for the load line, which gives Vce as a function of Ic and which you still refuse to do. Why won't you do it?

You have two points on that curve. Vce = 5 V when Ic = 0 mA and Vce = 0 V when Ic = 7.5 mA. You know that it is a system so you know it is a straight line.

ic = 20 * 43uA = 860 uA
vce = 10v - 1kohms * 860 uA = 9.14v

Think about this. Does it make sense? You have already determined that Vce when Ic is ZERO is 5 V and that if there is ANY Ic that the voltage goes down. So how can having an Ic of 0.86 mA now result in a Vce of 9.14 mA?

ic = β * 43uA
vce = 10v -1kohms * (β * 43uA ) so i left β as a parameter

This is the right idea, but you need to use the load line. Which means you need to finally come up with the equation for the load line.

 

You started off on the right track, but just because Ic is 4.3 mA does NOT mean that the current in the 1 kΩ resistor is 4.3 mA.

that's what i thought first that's why i named it by i1 but when we told me to assume β i followed the path of the loop and ignored the load line but now i thin about it there is no reason to ignore it and the equation is :
ic = 100 * 43 uA = = 4.3 mA
and vce = 10v - 1k * ( 4.3 mA + irl )
vce = 10v - 1000irl + 4.3v
and now what ??
this circuit is gonna make me mad i know there gotta be something that connects all these separated equations and thoughts together but i can't find it
i can't even think about my lessons i'm just thinking about it what's the thing i've been missing and every time i can't reach it

The whole point of developing the Thevenin equivalent as seen by the collector-emitter of the transistor is so that you could easily address this part of the problem. That is why I keep asking you to come with with the equation for the load line, which gives Vce as a function of Ic and which you still refuse to do. Why won't you do it?

i'm not refusing to do it i can't exactly find the equation i tried but didn't find one written equation and looks like this is my whole problem i can't find the equation u r pointing at
what kind of equation that will give me vce as function of ic FOR THE LOAD LINE ??
clearly that ic is the current flowing in the transistor and it's not equal to irl because i tried it before and failed so ic can't be i input for the load line even though vce is the voltage input for the load line so how can i come up with an equation for the load line that includes vce as a function of ic ??

Vce = 5 V when Ic = 0 mA and Vce = 0 V when Ic = 7.5 mA

but u didn't say that ic really equals 0mA when vce = 5v u told me that how could ic be 0 when ib = 43uA and vce = 5v and i said that i doubted this value and u told me then when u thin that some thing's strange u must stop and so i thought that
ic = 0mA is wrong

Think about this. Does it make sense? You have already determined that Vce when Ic is ZERO is 5 V and that if there is ANY Ic that the voltage goes down. So how can having an Ic of 0.86 mA now result in a Vce of 9.14 mA?

well it's not a big value for ic it's too small any way since u mean that ic= 0mA when vce = 5v is correct so it's a wrong value of course

This is the right idea, but you need to use the load line. Which means you need to finally come up with the equation for the load line.

and this is exactly what i can't come up with no matter how i try but now i don't have any thing right in the solution but just ib and ic sat and this makes me feel horrible i didn't have a trouble with any circuit for all this time
i started to doubt that my way of trying to find the solution is wrong from the very beginning shall i look for a new angle and try to start back again in a different way ?
even though i don't know how can i find a different way
too close to be solved yet too far

 

Let's take a step away from this problem entirely for a moment.

Let's say that I draw a line on a piece of graph paper. As usual, the x-axis is the horizontal axis and the y-axis is the vertical axis. I give you the following two points that are on the line.

<x,y> = <10, 0>
<x,y> = <0, 15>

Can you come up with values of m and b such that

y = mx + b

for this line?

 

i went back to the beginning and wrote the equations again
ib= ( 5-0.7 ) / 100000 = 43uA
ic = β * 43uA
irl =[ (vce)-(-5) ] / 2000
vce = 10v - (1000*i1)
ic = i1 + irl
i'm sure of these equations but can't find a number just parameters
and when it comes to thevenin theorem it gets complicated

 

i went back to the beginning and wrote the equations again
ib= ( 5-0.7 ) / 100000 = 43uA
ic = β * 43uA
irl =[ (vce)-(-5) ] / 2000
vce = 10v - (1000*i1)
ic = i1 + irl
i'm sure of these equations but can't find a number just parameters
and when it comes to thevenin theorem it gets complicated

I see that you still refuse to properly use units in even the simplest equations.

The Thevenin Theorem does NOT make it more complicated. It makes it considerably simpler.

Go back to the circuit I provided here:

https://forum.allaboutcircuits.com/threads/need-a-help-with-these-two-circuits.146559/#post-1248321

The whole point is that, as far as the transistor is concerned, that circuit is completely indistinquishable from the original circuit. So analyze that circuit.

 

Let's say that I draw a line on a piece of graph paper. As usual, the x-axis is the horizontal axis and the y-axis is the vertical axis. I give you the following two points that are on the line.

<x,y> = <10, 0>
<x,y> = <0, 15>

Can you come up with values of m and b such that

y = mx + b

for this line?

yes it's easy [ even though i didn't understand what's the relationship between this and the circuit ] :
y = m * x + b
0 = 10m + b
15 = 0 + b
b = 15
0 = 10m + 15
10m = -15
m = -1.5
do u mean that y is vce and x is ic ? then if it's the load line equation :
vce = 2k ic +(-5)
but that means ic is irl !!
if i keep going then i have : when vce = 0 then
0 = 2k ic = (-5)
ic = (-5) / 2000
ic = 2,5 mA
and when vce = 15 then
15 = 2k ic - (-5)
10 = 2k ic
ic = 5 mA
now if i try vce = 5v , ic = 0mA and vce = 0v ic = 7.5mA then
0 = 2k * ic - (-5)
ic = (-5) / 2k
ic = 2.5mA
and now i can't go any further it's supposed to be that when vce = 0v ic will be .5mA
and this proves that ic isn't irl SO HOW ???

 

yes it's easy [ even though i didn't understand what's the relationship between this and the circuit ] :

I suspect you are what is known as "strongly field dependent". This is neither good nor bad, but just one aspect of cognitive learning styles.

Having said that, strongly field dependent learners often have trouble breaking apart problems and looking at the constituent parts in isolation. They also tend to have trouble taking what they have learned from one problem and applying it to others. The reason is that their understanding of each problem is strongly influenced by the context of the problem. Change the context, and it simply seems like a completely unrelated problem.

Most scientists and engineers are strongly field independent, which is the opposite. We can easily break problems down and apply what we learn about seemingly unrelated fields of study to the problem at hand. But strongly field independent folks tend to have a hard time keeping the big picture in mind, whereas strongly field dependent folks do that very naturally.

do u mean that y is vce and x is ic ? then if it's the load line equation :
vce = 2k ic +(-5)

Yes, that is the idea, but the load line is dictated by the Thevenin equivalent.

You can either come up with the equation of the line for Vce (y) as a function of Ic (x) or for Ic (y) as a function of Vce (x).

Since we have Ic (because we have Ib and can either assume a value of beta or just leave it as a parameter), it makes sense to make that our x-axis parameter.

but that means ic is irl !!

Good catch. So you know it is wrong. No point going any further until you resolve this because you KNOW that ic is NOT irl (they might be equal at some carefully chosen operating point, but they are still two different currents).

So what is wrong? You are using only PART of the circuit that connects to the collector-emitter of the transistor. You need to use the entire circuit that does so. But you have a very, very, very simple equivalent for that entire circuit. So use it!

if i keep going

DON'T KEEP GOING!

You KNOW that everything you do from this point on is worthless! Why waste the time and effort only to end up with a result that you know is wrong?

Go back to determining the equation for the load line.

I have a curve that I know is a straight line. I know two points on that line, namely:

<Ic, Vce> = <0 mA, 5 V>
<Ic, Vce> = <7.5 mA, 0 V>

What is the equation for the line

Vce = R⋅Ic + Vce0

This is just using different names for y = mx + b.

 

I see that you still refuse to properly use units in even the simplest equations.

i'm sorry i forgot but i can't think of anything but the solution so i forget lots of things

The Thevenin Theorem does NOT make it more complicated. It makes it considerably simpler.

this is what it's supposed to do but i feel it made this circuit more complicated

The whole point is that, as far as the transistor is concerned, that circuit is completely indistinquishable from the original circuit. So analyze that circuit.

i did it many many times
vce = vthev - (req * ic )
vce = 5 - ( 7.5mA * 0.667k )
vce = 5-5 = 0v and this has actually been proved so what's wrong with this ??

 

i'm sorry i forgot but i can't think of anything but the solution so i forget lots of things

this is what it's supposed to do but i feel it made this circuit more complicated

i did it many many times
vce = vthev - (req * ic )

vce = vthev - (req * ic )

THIS is your load line equation (though it would be nice if you were to discover that there is a key called "Shift" that would allow you to write this more appropriately as:

Vce = Vthev - (Req * Ic )

vce = 5 - ( 7.5mA * 0.667k )
vce = 5-5 = 0v and this has actually been proved so what's wrong with this ??

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.

 

I suspect you are what is known as "strongly field dependent". This is neither good nor bad, but just one aspect of cognitive learning styles

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

Vce = R⋅Ic + Vce0

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