# Homework Help: Basic electronics problem with an amplification factor for a transistor

1. Aug 18, 2011

### Femme_physics

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

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

1. The problem statement, all variables and given/known data
http://img600.imageshack.us/img600/7950/back1j.jpg [Broken]

http://img690.imageshack.us/img690/999/back2wzt.jpg [Broken]

2. Relevant equations

http://img27.imageshack.us/img27/3748/back3za.jpg [Broken]

3. 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 = IC/IB

Plugging in to the second formula:

IE = IB (1+beta)

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

IE = IB + IC

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?

Last edited by a moderator: May 5, 2017
2. Aug 18, 2011

### a-tom-ic

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

I think you need to know the value of the supply voltage, else you cant 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:
$$I_B = U_{R_B}/R_{B}$$

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 $U_{BE}$.

Last edited: Aug 18, 2011
3. Aug 18, 2011

### skeptic2

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

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?

4. Aug 18, 2011

### Femme_physics

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

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

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!

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

5. Aug 18, 2011

### a-tom-ic

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

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.

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

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.

6. Aug 18, 2011

### Femme_physics

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

Oh, I added the direction of the emitter.

http://img26.imageshack.us/img26/4273/directd.jpg [Broken]
Sorry.
Do all transistors have the same number of pins? three?

Thanks a lot for that :)

Well, first off you asked for it so

http://img29.imageshack.us/img29/6646/solsol.jpg [Broken]

How'd you figure all that out without calculations?

Last edited by a moderator: May 5, 2017
7. Aug 18, 2011

### Femme_physics

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

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?

8. Aug 18, 2011

### a-tom-ic

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

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 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?

I do not understand what you mean here?

Last edited by a moderator: May 5, 2017
9. Aug 18, 2011

### a-tom-ic

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

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

10. Aug 18, 2011

### Femme_physics

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

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?

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

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

11. Aug 18, 2011

### a-tom-ic

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

See attached picture.

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 im back =P

#### Attached Files:

• ###### Transistor_Exercise.jpg
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12. Aug 18, 2011

### Femme_physics

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

I see. Thanks^^

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 wanna 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?

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

13. Aug 18, 2011

### Ouabache

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

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.

yes

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".

Last edited by a moderator: Apr 26, 2017
14. Aug 18, 2011

### a-tom-ic

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

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.

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.

#### Attached Files:

• ###### Transistor_Exercise2.JPG
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15. Aug 20, 2011

### Femme_physics

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

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...

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 [Broken]

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

http://img818.imageshack.us/img818/6084/isok.jpg [Broken]

Last edited by a moderator: May 5, 2017
16. Aug 20, 2011

### I like Serena

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

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.

17. Aug 20, 2011

### a-tom-ic

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

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?

18. Aug 20, 2011

### Femme_physics

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

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 [Broken]

http://img856.imageshack.us/img856/5305/twooooooo.jpg [Broken]

http://img27.imageshack.us/img27/3748/back3za.jpg [Broken]

Ah but here the base is connected, 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.

Last edited by a moderator: May 5, 2017
19. Aug 20, 2011

### I like Serena

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

Yep.

Yes.

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. )

Last edited: Aug 20, 2011
20. Aug 20, 2011

### a-tom-ic

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

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.)

21. Aug 20, 2011

### Ouabache

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

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.

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,
more easily

Last edited: Aug 20, 2011
22. Aug 21, 2011

### Femme_physics

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

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 [Broken]

http://img7.imageshack.us/img7/1569/mul2h.jpg [Broken]

http://img508.imageshack.us/img508/1609/mul3.jpg [Broken]

http://img716.imageshack.us/img716/2519/mul4.jpg [Broken]

Atomic? ILS?

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

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

Yes, seems easy! Just a lot of algebra.

Last edited by a moderator: May 5, 2017
23. Aug 21, 2011

### I like Serena

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

Hey Fp!

In your problem statement you have:
VBE = 0.7 [V]
RB = 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).

And I don't understand what you did in your third picture.

Last edited: Aug 21, 2011
24. Aug 21, 2011

### I like Serena

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

Suppose you had a mechanics problem with this moment sum:

Σ M = 0; -FB rB - MBE + 12 [N m] = 0

and you would know that rB = 100 [mm] and MBE = 0.7 [N m].

How would you solve it?

25. Aug 21, 2011

### a-tom-ic

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

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

My last picture in attachment.

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