Electronics with Transistor (another problem)

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Homework Help Overview

The discussion revolves around a problem involving a transistor circuit, specifically focusing on finding the collector current (Ic), emitter current (Ie), and base current (Ib). Participants are analyzing the implications of certain conditions, such as Ic being zero, and discussing the calculations related to the transistor's operation modes.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the correctness of their calculations for Ic, Ie, and Ib, with some questioning the implications of defining Ic as zero. There are inquiries about the relevance of using scientific notation for current values and concerns about potential confusion with units. Others explore the definitions of saturation and cutoff points in transistor operation.

Discussion Status

Some participants have provided guidance on the calculations and the interpretation of the problem, while others are still seeking clarity on specific aspects of the circuit and the definitions involved. Multiple interpretations of the problem's requirements are being explored, particularly regarding the conditions under which Ic can be considered zero.

Contextual Notes

Participants note that there are words in the problem statement that are difficult to translate, which may affect their understanding. The problem includes specific conditions for finding Vce and Ic at different operational points of the transistor.

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

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?
 
Last edited by a moderator:
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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
 
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 :smile:


Anyway, I solved it and will post the full scan later^^ thanks.
 
It is all right. But engineers do not like to write out so many zeros.

ehild
 
Good mooorning Fp! :smile:

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.
 
Last edited:
Goo morning! :smile:

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.
 
You just calculated Ic to be 1.599 mA.
And beta was given as 30.
So what did you do?
 
)
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?
 
  • #10
Btw, since you mentioned I should have a stamp, I've got one now! :wink:

However, I'm still waiting for a solution to put it on...
 
  • #11
Femme_physics said:
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?
 
  • #12
I like Serena said:
Btw, since you mentioned I should have a stamp, I've got one now! :wink:

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"
 
  • #13
I like Serena said:
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!
 
  • #14
Femme_physics said:
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.
 
  • #15
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?
 
  • #16
Femme_physics said:
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
 
  • #17
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 :smile: so my answers are correct?
 
  • #18
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.
 
  • #19
Femme_physics said:
Aha :smile: so my answers are correct?

Answers? :confused:

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. :smile:
 
  • #20
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. :smile:
[/quote]

Where'd u come up with that?
 
  • #21
I've just switched to the main branch of my office. At least there I can read a scan without too much hassle! :smile:

Femme_physics said:
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.
 
  • #22
Btw, the formula Ie=Ib+Ic also still works.
 
  • #23
I've just switched to the main branch of my office. At least there I can read a scan without too much hassle!

:biggrin: 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?
 
  • #24
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?
 
  • #25
Femme_physics said:
:biggrin: where is that?

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


Femme_physics said:
:And is Ib still equal to the same thing?

Nope.
But it doesn't really matter what Ib is I think.
 
  • #26
Femme_physics said:
But what about the other stuff? I1? I2? They get to stay the same?

Huh? :confused:

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

Afaik there is no I1 or I2 in this problem.
 
  • #27
Aaaaaand I'm off to breakfast! :rolleyes:
 
  • #28
I like Serena said:
Huh? :confused:

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 :)
 
  • #29
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 :smile:
 
  • #30
Femme_physics said:
My bad, I was mixing up problems lol

]

beteavon :)

Rav todot! :smile:
 

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