Electronics with Transistor (another problem)

[Femme_physics]

Sorry I'm jumping from one problem to another, I just have plenty of time today and I got my scanner

Homework Statement

Find Ic, Ie, Ib

(circuit drawing and data in the solution)

The Attempt at a Solution

http://img263.imageshack.us/img263/3285/findcurrents.jpg

Is that correct for Ie and Ib?

 

[ehild]

It is correct, but use scientific notations or mA units instead of the lot of zeros.
Note that kΩ multiplied by mA is also Volt. If you plug-in resistances in kΩ units, you get the currents in mA-s.

ehild

 

[Femme_physics]

It is correct, but use scientific notations or mA units instead of the lot of zeros.

Yea but does it really matter? I mean, either way it's correct. I'll still get full points for the right score.

Note that kΩ multiplied by mA is also Volt. If you plug-in resistances in kΩ units, you get the currents in mA-s.

True. I just sometimes worried I'll get confused with the milis and the kilos and whatever, so I stick to the originals


Anyway, I solved it and will post the full scan later^^ thanks.

 

[ehild]

It is all right. But engineers do not like to write out so many zeros.

ehild

 

[Femme_physics]

http://img534.imageshack.us/img534/2769/fixiation1.jpg

http://img148.imageshack.us/img148/3091/fixiation2.jpg

http://img7.imageshack.us/img7/6936/fixiation3.jpg


I'm just not sure about the last 2 clauses...is it all correct?

 

[I like Serena]

Good mooorning Fp!

All your calculations are correct.
Except the last 2 clauses.

You seem to have left out part of the problem when you stated that Ic=0...?

Anyway, if Ic is 0, you obviously have a different circuit, and then you cannot assume that Ib is unchanged.

 

[Femme_physics]

Goo morning!

You seem to have left out part of the problem when you stated that Ic=0...?

Anyway, if Ic is 0, you obviously have a different circuit, and then you cannot assume that Ib is unchanged.

How come? when I typed beta = Ic/Ib

I defined Ic as 0.

 

[I like Serena]

You just calculated Ic to be 1.599 mA.
And beta was given as 30.
So what did you do?

 

[Femme_physics]

)
You just calculated Ic to be 1.599 mA.
And beta was given as 30.
So what did you do?

Yea that was in the old calculations, but then they have redefined IC = 0 so I had to adapt?

 

[I like Serena]

Btw, since you mentioned I should have a stamp, I've got one now!

However, I'm still waiting for a solution to put it on...

 

[I like Serena]

Yea that was in the old calculations, but then they have redefined IC = 0 so I had to adapt?

They can't *just* redefine that!
But what I suspect, is that the battery on the right should be gone.
Did you add that yourself?

 

[Femme_physics]

Btw, since you mentioned I should have a stamp, I've got one now!

However, I'm still waiting for a solution to put it on...

I still don't see what's wrong? Oh and there are words that I can't translate in the question. that's why there's a little gap in the question. I'll try though.

They said "Find Vce at the section point (when Ic = 0)"

And the last question is to "find Ic at the saturation point when Vce = 0"

 

[Femme_physics]

They can't *just* redefine that!
But what I suspect, is that the battery on the right should be gone.
Did you add that yourself?

No, that's what the question says...really!

 

[I like Serena]

No, that's what the question says...really!

What does the question say exactly?


As I see it there are only 2 ways for Ic to be 0.

One if you remove the voltage on the collector, so no current can flow.

And the other possibility is if you remove the voltage on the base, so the transistor behaves like an open switch.

 

[Femme_physics]

What does the question say exactly?

I wrote that in 2 posts above u

They said "Find Vce at the section point (when Ic = 0)"

And the last question is to "find Ic at the saturation point when Vce = 0"

I suppose it could just be one of those questions where you can give a literal answer without calculations. You know, there is such thing as easy questions with a simple one-liner answer?

 

[I like Serena]

I still don't see what's wrong? Oh and there are words that I can't translate in the question. that's why there's a little gap in the question. I'll try though.

They said "Find Vce at the section point (when Ic = 0)"

And the last question is to "find Ic at the saturation point when Vce = 0"

Aha!
A transistor has several modes of operation.

Apparently the question asks something about the "Saturation" and "Cut-off" points of the transistor.

Saturated means the base current is so high that the "transistor switch" is fully closed, meaning Vce is 0.

Cut-off means the base voltage is so low, that the "switch" is fully open, and Ic is 0.

The

 

[Femme_physics]

Saturated means the base current is so high that the "transistor switch" is fully closed, meaning Vce is 0.

Cut-off means the base voltage is so low, that the "switch" is fully open, and Ic is 0.

Aha so my answers are correct?

 

[I like Serena]

I think they want you to redo the calculations, for both cases (which will be much easier).

Saturation is equivalent to removing resistor Rb, and then calculate Ic.

Cut-off is equivalent to removing the battery on the left, and calculate Vce.

 

[I like Serena]

Aha so my answers are correct?

Answers?

Afaik you only did Ic=0, which is the cutoff.
You can't calculate beta like you did, because Ib would be zero too.
But yes, in Cutoff, Vce would be 12 V.

 

[Femme_physics]

You can't calculate beta like you did, because Ib would be zero too.

so I'm not allowed to use the transistor formula in this case?

Then what can I?

But yes, in Cutoff, Vce would be 12 V.

[/quote]

Where'd u come up with that?

 

[I like Serena]

I've just switched to the main branch of my office. At least there I can read a scan without too much hassle!

so I'm not allowed to use the transistor formula in this case?

Then what can I?

Where'd u come up with that?

The formula Ic = beta x Ib only works during normal ("forward-biased") operation of the transistor.

But in the cutoff case you can still use KVL on the right hand side loop with Ic=0, which is basically what you did.

In the saturated case, again you should apply KVL on the right hand side loop with Vce=0 and an unknown Ic.

 

[I like Serena]

Btw, the formula Ie=Ib+Ic also still works.

 

[Femme_physics]

I've just switched to the main branch of my office. At least there I can read a scan without too much hassle!

where is that?

The formula Ic = beta x Ib only works during normal ("forward-biased") operation of the transistor.

Ahhh...

But in the cutoff case you can still use KVL on the right hand side loop with Ic=0, which is basically what you did.

In the saturated case, again you should apply KVL on the right hand side loop with Vce=0 and an unknown Ic.

Right, give me a moment...

Btw, the formula Ie=Ib+Ic also still works.

And is Ib still equal to the same thing?

 

[Femme_physics]

But in the cutoff case you can still use KVL on the right hand side loop with Ic=0, which is basically what you did.

In the saturated case, again you should apply KVL on the right hand side loop with Vce=0 and an unknown Ic.

But what about the other stuff? I1? I2? They get to stay the same?

 

[I like Serena]

where is that?

I've just SMS-ed you a picture of my main branch!

:And is Ib still equal to the same thing?

Nope.
But it doesn't really matter what Ib is I think.

 

[I like Serena]

But what about the other stuff? I1? I2? They get to stay the same?

Huh?

Aren't you mixing up problems here?
(Or am I )

Afaik there is no I1 or I2 in this problem.

 

[I like Serena]

Aaaaaand I'm off to breakfast!

 

[Femme_physics]

Huh?

Aren't you mixing up problems here?
(Or am I )

Afaik there is no I1 or I2 in this problem.

My bad, I was mixing up problems lol

Aaaaaand I'm off to breakfast!

]

beteavon :)

 

[Femme_physics]

Anyway ignore the I1 I2 thing I made a mistake so I got to redo the exercise anyway...I wasn't even sure what I was asking you there. I'll straighten up my act

 

[I like Serena]

My bad, I was mixing up problems lol

]

beteavon :)

Rav todot!

 

[I like Serena]

I found this little electronics store... and they could fix my charger!
Me and my laptop are back in business again! Yeah! :!)

 

[Ouabache]

(from your post #5)
I'm just not sure about the last 2 clauses...is it all correct?

Hi FP !

In working through your exercise, I agree with your sol'ns for Ib, Ie, Vce, Vre. Good job !
However I don't believe you've correctly solved for Vb. Why do you feel Vb (the voltage at the base of the transistor) is only equal to IbRb??
Hint: if 12V is the potential (metaphorically, at the top of the hill) and you come part way down the hill to Vb. What should Vb equal?

I agree with ehild that answers are preferred in engineering, w/o so many zeros. (either scientific notation or using prefixes mA , $\mu$A, mV, $\mu$V). If you really like the zeros (for your peace of mind), you might state your answer e.g. 0.000043A = 0.043mA

There is something in this problem that will help you on the https://www.physicsforums.com/showthread.php?t=525613" (the one with the voltage divider R1 and R2, biasing network). In fact the labeling of the variables (eg. Vb, Rb) in the current problem is a hint at how useful it is to define them this way.

What I am referring to is your equation in the current problem that looks something like:
(a) : Vb - IbRb - Vbe - IeRe = 0
I recommend using this equation, in your earlier exercise. I promise, it will simplify your calculations immensely. (You may want to review how I defined Vb and Rb in https://www.physicsforums.com/showthread.php?t=525613&page=3,#35")

I miss your hand drawn-diagrams (photographed). Some of your new 'scans' are not bold or enlarged enough to read clearly. Your hand-drawn photographed images added a wonderful creative touch to your posts