Basic electronics problem with an amplification factor for a transistor

[Femme_physics]

Basic electronics problem with an "amplification factor for a transistor"

((I think it still counts as "introductory physics" -- let me know if I'm wrong!))

Homework Statement

http://img600.imageshack.us/img600/7950/back1j.jpg

http://img690.imageshack.us/img690/999/back2wzt.jpg

Homework Equations

http://img27.imageshack.us/img27/3748/back3za.jpg

The Attempt at a Solution

So "beta" means "amplification factor for a transistor" - I got it written down but I'm not sure what it means in theory. But, I do see the formulas, so I will try to play with them

Well, clearly the formulas are not enough to help me through it. Plugging it to the first formula:

60 = I_C/I_B

Plugging into the second formula:

I_E = I_B (1+beta)

And to the last formula there's nothing to plug, it's just

I_E = I_B + I_C

Soooooo...

Now I got two other tools I know, KCL an KVL,

But I can't use any of them since this is NOT a closed circuit! In fact, I'm not sure what these arrows up mean. Can anyone help me please?

 

[a-tom-ic]

I think you need to know the value of the supply voltage, else you can't solve it in numbers but only with the supply voltage as a parameter. Let's say "U" is the supply voltage:

So you can calculate the current $I_B$ with:
$$U_{R_B} = U - 0,7 V$$
Ohm's Law:
$$I_B = U_{R_B}/R_{B}$$
and then advance with your formulas to get the other results.

For your question about the arrows. They just tell you how your voltage was defined. If you turn the arrow direction you need to change the sign of the voltage. And its up to you how you define them. However with $U_{BE}$ is equal to 0,7 V the arrow should point in the other direction in my opinion.
I have done that already with $U_{BE}$.

 

[skeptic2]

Yes, beta is the current gain of the transistor - not necessarily the voltage gain. The first thing to do is calculate the Ib. Note that Ib is the current that flows from the up arrow on the left, through RB and from base to the emitter. Of course to calculate the current you need to know the supply voltage which you haven't included. If you call the supply voltage Vs, can you write an expression for Ib using RB and Vbe?

 

[Femme_physics]

Yes, beta is the current gain of the transistor - not necessarily the voltage gain.

You mean the voltage drop? Just like resistors have voltage drops, so does transistors?

to the emitter.

I can't find a good definition of emitter and I've been googling like crazy. wiki leads me here

http://en.wikipedia.org/wiki/Emitter

And I don't know which one to pick since there are 5 links under "electronics". And no, I can't find it in my lecture notes or notebook from some reason. Ugh!

I think you need to know the value of the supply voltage, else you can't solve it in numbers but only with the supply voltage as a parameter. Let's say "U" is the supply voltage:

In the solution to this problem I see our lecturer solved it using a full result of Ib in miliampers without any unknowns.

 

[a-tom-ic]

You mean the voltage drop? Just like resistors have voltage drops, so does transistors?

There will be a certain voltage difference between each of the pins of the transistor. Nothing was said about that with the current gain definition. It just means that for a base current Ib= 1mA and a current gain beta = 100, your collector current Ic will be 100 mA.

I can't find a good definition of emitter and I've been googling like crazy. wiki leads me ...

Emitter is a pin of your transistor: http://www.technologystudent.com/elec1/transis1.htm
Your picture in your initial post, lacks the arrow to direction of the transistor symbol. For all symbols see: http://en.wikipedia.org/wiki/Transistor

In the solution to this problem I see our lecturer solved it using a full result of Ib in miliampers without any unknowns.

Can you show us the solution then? Or is it just the number without any other information.
Skeptic2 as well as i assumed that you need the value of the supply voltage. If your teacher has that voltage of course he can give an exact answer. Maybe he assumes TTL Logic with supply voltage of 5 V for example. Try 5V and look if your teachers result is the same as yours then.

 

[Femme_physics]

Oh, I added the direction of the emitter.

http://img26.imageshack.us/img26/4273/directd.jpg
Sorry.

There will be a certain voltage difference between each of the pins of the transistor.

Do all transistors have the same number of pins? three?

Emitter is a pin of your transistor: http://www.technologystudent.com/elec1/transis1.htm
Your picture in your initial post, lacks the arrow to direction of the transistor symbol. For all symbols see: http://en.wikipedia.org/wiki/Transistor

Thanks a lot for that :)

Can you show us the solution then? Or is it just the number without any other information.
Skeptic2 as well as i assumed that you need the value of the supply voltage. If your teacher has that voltage of course he can give an exact answer. Maybe he assumes TTL Logic with supply voltage of 5 V for example. Try 5V and look if your teachers result is the same as yours then.

Well, first off you asked for it so

http://img29.imageshack.us/img29/6646/solsol.jpg

There will be a certain voltage difference between each of the pins of the transistor. Nothing was said about that with the current gain definition. It just means that for a base current Ib= 1mA and a current gain beta = 100, your collector current Ic will be 100 mA.

How'd you figure all that out without calculations?

 

[Femme_physics]

And how am I supposed to know from the drawing which leg is the base, which leg is the emitter, and which leg is the collector?

 

[a-tom-ic]

Do all transistors have the same number of pins? three?

There exist packages of transistors with more pins. Some have an additional pin for the chasis additionally to the regular ones for base, collector, emitter. Special "transistors" have a second base input. (Basically 2 transistors with separate base in the same package). Maybe someone with more knowledge can answer you this question in more detail.

Well, first off you asked for it so
http://img29.imageshack.us/img29/6646/solsol.jpg

Well the formulas given in that picture are exactly the ones i suggested in my first reply to your question. If you tell us know by which magic your teacher came to the 12 V, you have your answer :). That is the supply voltage he assumed. Btw. you made an writing error its 100kOhm (100 000 Ohm) not 120kOhm in your initial picture, so 0,113 mA would be correct?

How'd you figure all that out without calculations?

I do not understand what you mean here?

 

[a-tom-ic]

And how am I supposed to know from the drawing which leg is the base, which leg is the emitter, and which leg is the collector?

The transistor symbol drawing defines which leg is which of the three.
Have a look again at all of the symbols of the wikilink given: Figure name "BJT and JFET symbols" http://en.wikipedia.org/wiki/Transistor

 

[Femme_physics]

Well the formulas given in that picture are exactly the ones i suggested in my first reply to your question. If you tell us know by which magic your teacher came to the 12 V, you have your answer :). That is the supply voltage he assumed. Btw. you made an writing error its 100kOhm (100 000 Ohm) not 120kOhm in your initial picture, so 0,113 mA would be correct?

In my initial picture I wrote 100k ohms I see. Where did I write 120k ohms? It can't be that my zero looks like a two?

I do not understand what you mean here?

Ib wasn't given to you. How did you decide it's 100?

There exist packages of transistors with more pins. Some have an additional pin for the chasis additionally to the regular ones for base, collector, emitter. Special "transistors" have a second base input. (Basically 2 transistors with separate base in the same package). Maybe someone with more knowledge can answer you this question in more detail.

It's ok, since I assume that for simple purposes and a introductory course we just use the 3 legged one then :) Thanks.

 

[a-tom-ic]

In my initial picture I wrote 100k ohms I see. Where did I write 120k ohms? It can't be that my zero looks like a two?

See attached picture.

Ib wasn't given to you. How did you decide it's 100?

The value 100 was an example for the current amplification factor B or beta however you call it. It was just an example. In YOUR case you used 60.

Im out now. Other questions for other people or when I am back =P

 

[Femme_physics]

The value 100 was an example for the current amplification factor B or beta however you call it. It was just an example. In YOUR case you used 60.

I see. Thanks^^

See attached picture

Maybe my teacher made an error, but I guess the odds are the I miscopied or something. At any rate, let's presume for our purposes it's actually 120k ohms and proceed from that PoV. I'm going to ignore the solution I posted since I can't understand it nor do I want to understand it - I want to try do my own solution (with of course help from you guys )Can I ask before I continue: what does an "amplification factor for a transistor" really mean? From what I understand it's how much does the transistor amplify the current? Could that be it?

If you tell us know by which magic your teacher came to the 12 V, you have your answer :).

But we don't even have a voltage source in the sketch. How am I supposed to know where it's located?

 

[Ouabache]

At any rate, let's presume for our purposes it's actually 120k ohms and proceed from that PoV. I'm going to ignore the solution I posted since I can't understand it nor do I want to understand it - I want to try do my own solution (with of course help from you guys )

I wouldn't ignore all the information just yet. That implied 12v source voltage (Vs) is going to be very useful. You will be able to use KVL easily with that source (remember your circuit does not need to look like a loop in order to apply KVL. In this circuit, it is implied that the circuit's ground and Vs (voltage source) ground are connected which does make a loop (i.e. from Vs through a path in your transistor circuit to ground and back to Vs).

If you presume Rb= 120K, you will just obtain a different base current than you have in your given solution (and since you calculate your collector Ic and emitter Ie currents from Ib, those will also be different). Once you understand how to determine find Ib, you may want to try this problem with both values; Rb=100K and Rb=120K, and see how that effects Ic & Ie.

Can I ask before I continue: what does an "amplification factor for a transistor" really mean? From what I understand it's how much does the transistor amplify the current? Could that be it?

yes

But we don't even have a voltage source in the sketch. How am I supposed to know where it's located?

You may want to skip ahead in your text and see what BJT (bipolar junction transistor) circuits look like. For now, they are keeping it simple and only considering DC applications. If you cannot find any good examples, here is one that indicates what you are asking about: http://www.ittc.ku.edu/~jstiles/312/handouts/Example DC Analysis of a BJT Circuit.pdf".

 

[a-tom-ic]

But we don't even have a voltage source in the sketch. How am I supposed to know where it's located?

The thing is, engineers are a) lazy, b) do not want to clutter their schematic drawing. Therefore the arrow is the equivalent to a voltage source. See attached picture part 1!
Part 2 is the same circuit explicitly redrawn with the voltage sources, so it should be more clear that there are voltage sources, and that the circuits are closed. Additionally i hope you now see where you should use the KVL.

Can I ask before I continue: what does an "amplification factor for a transistor" really mean? From what I understand it's how much does the transistor amplify the current? Could that be it?

If we are talking again about the current gain $B=\frac{I_C}{I_B}$ then B is just the proportional constant which approximatly relates the base current to the collector current.

 

[Femme_physics]

I wouldn't ignore all the information just yet. That implied 12v source voltage (Vs) is going to be very useful.

Wait, so it's GIVEN to me that the source voltage is 12V?

---

And actually, it really should be 100k ohms, there is a mistake (or a typo more correctly) in the solution I posted of my teacher. I showed it to him yesterday and we were both kinda surprised nobody noticed it and everyone just copied from the board. Heh. Anyway, back to business...

The thing is, engineers are a) lazy, b) do not want to clutter their schematic drawing. Therefore the arrow is the equivalent to a voltage source. See attached picture part 1!

Ahh I see :)

Well, anyway, this is how my lecturer drew it for me yesterday when I asked him.

http://img21.imageshack.us/img21/9081/lecturer.jpg

So I'm going to take it from there. And this is my idea of what IC equals to...


http://img818.imageshack.us/img818/6084/isok.jpg

 

[I like Serena]

Hey Fp!

I'm afraid the transistor throws in a resistance between C and E.
So your voltage law won't work like this.

A transistor is a bit weird thingy.
If its base B is not connected, it throws up an infinite resistance block between C and E.
But if there is only a very small current applied to its base B, the points C and E start behaving as if they are connected.

 

[a-tom-ic]

You ignored that there is a voltage drop between the collector and emitter pin of the transistor (you called that "Vce"). Vce is missing in your KVL equation in the last picture. Therefore your solution is wrong.

Since i assume you have your bias current calculated already, why not just use the equations you even have given us in your first post?

 

[Femme_physics]

Since i assume you have your bias current calculated already, why not just use the equations you even have given us in your first post?

Ahhh. I see

So did I build all my equations correct now and I can start using them to solve the problem?

http://img694.imageshack.us/img694/7/oneeee.jpg

http://img856.imageshack.us/img856/5305/twooooooo.jpg

Uploaded with ImageShack.us
http://img27.imageshack.us/img27/3748/back3za.jpg

A transistor is a bit weird thingy.
If its base B is not connected, it throws up an infinite resistance block between C and E.

Ah but here the base is connected, yes?

But if there is only a very small current applied to its base B, the points C and E start behaving as if they are connected.

Interesting, and noted. Why does this happen? I remember that coils are not actually touching each other, but "behave as though they're connected" when a current pass through them.

 

[I like Serena]

Ahhh. I see

So did I build all my equations correct now and I can start using them to solve the problem?

Yep.

Ah but here the base is connected, yes?

Yes.

Interesting, and noted. Why does this happen? I remember that coils are not actually touching each other, but "behave as though they're connected" when a current pass through them.

A current in a coil generate a magnetic field, and a magnetic field induces a current in a coil.
(When did you learn about coils?)

A transistor consists of 2 materials connected to each other.
Left to themselves they block the current (between collector and emitter).
But if a couple of electrons are drawn away from the place where they connect (where the base B is connected), suddenly current can pass through.
(I don't know why this is. Something with the strange material properties of Silicium. )

 

[a-tom-ic]

So did I build all my equations correct now and I can start using them to solve the problem?

Both KVLs are correct, use them! (However you will not need the one for the output for the transistor. Only if you want to calculate Vce too it will be useful.)

 

[Ouabache]

Wait, so it's GIVEN to me that the source voltage is 12V?

כן ;when you realize the problem cannot be solved numerically without a
source voltage (Vs), then using the solution was a useful way to deduce the value of
Vs.

And actually, it really should be 100k ohms, there is a mistake (or a typo more correctly) in the solution I posted of my teacher. I showed it to him yesterday and we were both kinda surprised nobody noticed it and everyone just copied from the board. Heh. Anyway, back to business...

Good job.. Given the equation you wrote down
Ib=11.3/12000 = 0.113[mA] , if you were to back calculate Rb as a double
check Rb = 11.3/0.113mA = ? (100K$\Omega$). You would note 100K$\Omega$ is
consistent with your original given value for Rb.
That is how I (and I believe a-tom-ic) both realized there must have been a typo there.
(ha-ha, where does typo come from? It comes from ancient times
when people used typewriters. typo=typographical error).
But as you pointed out, it was your lecturer's written error.

Did you happen to look at the BJT DC circuit reference I gave you?
They describe the same type of problem you are working on.
They may have thrown in a couple more resistors and the applied voltages
may look slightly different but I wouldn't let that throw you.

Their procedure is accurate and straight forward and I believe,
will help you understand how to approach these transistor circuits
more easily

 

[Femme_physics]

Just flunk at too many unknowns with too many equation. I guess over than 4 unknowns and I get confused.

http://img269.imageshack.us/img269/7353/mul1x.jpg http://img7.imageshack.us/img7/1569/mul2h.jpg

http://img508.imageshack.us/img508/1609/mul3.jpg http://img716.imageshack.us/img716/2519/mul4.jpg Atomic? ILS?

Good job.. Given the equation you wrote down
Ib=11.3/12000 = 0.113[mA] , if you were to back calculate Rb as a double
check Rb = 11.3/0.113mA = ? (100KΩ). You would note 100KΩ is
consistent with your original given value for Rb.
That is how I (and I believe a-tom-ic) both realized there must have been a typo there.
(ha-ha, where does typo come from? It comes from ancient times
when people used typewriters. typo=typographical error).
But as you pointed out, it was your lecturer's written error.

Hah, true, and yes. Thanks for that historical tid-bit

Did you happen to look at the BJT DC circuit reference I gave you?
They describe the same type of problem you are working on.
They may have thrown in a couple more resistors and the applied voltages
may look slightly different but I wouldn't let that throw you.

Alright. Well, one step at a time. I want to see I can solve this first

Their procedure is accurate and straight forward and I believe,
will help you understand how to approach these transistor circuits
more easily

Yes, seems easy! Just a lot of algebra.

 

[I like Serena]

Hey Fp!

In your problem statement you have:
V_BE = 0.7 [V]
R_B = 100 kOhm
etcetera

Do you have an equation with only 1 unknown in it?
Btw, I think the reason that you got stuck, is because you used the same equation twice.
Your 4th equation was derived from the 3rd and the 5th (in your first picture).

Furthermore you dropped a minus sign in your second picture.
And I don't understand what you did in your third picture.

 

[I like Serena]

Suppose you had a mechanics problem with this moment sum:

Σ M = 0; -F_B r_B - M_BE + 12 [N m] = 0

and you would know that r_B = 100 [mm] and M_BE = 0.7 [N m].How would you solve it?

 

[a-tom-ic]

Im sorry feme that i might complain now, but I am a bit worried that you do not read what we write?

My last picture in attachment.

 

[Femme_physics]

Well, it may take me a bit of time to reply occasionally but I'm trying my best

I forgot that Vbe was given to me! Thanks for pointing it out ILS

And thanks atomic, yes, someone I miss things but thanks for handing it to me in a silver platter. I'll try to read harder between the lines next time^^

Now it's easy! LOL real easy!

Hah ILS thanks for the mechanics correlation but I got it

http://img810.imageshack.us/img810/4350/69302652.jpg

Uploaded with ImageShack.us

Now I'll get to other stuff but if I get into problems I'll let you know

Thank you!