Transistors circuit - Basic questions

[I like Serena]

I suppose it's to ask my teacher. Should I basically ask if we're allowed to use voltage dividers? Or "on what circumstances are we allowed to use them"?

Ask: "Can we use voltage dividers when current flows to the base of a transistor from the middle?"

If bigger currents could flow, you should not use voltage dividers.
If no significant current flows (for instance to an op-amp) you can use voltage dividers.

It is certainly valid to verify your results, which you should always do.

Yep, hence here :)

Good!

 

[ehild]

Wait let me try it ... I had it before I think:

http://img252.imageshack.us/img252/1037/vbez.jpg

Right?

It looks right! Substitute IC=0.8 mA and VBE=0.6 V, cancel I2 from the first two equations and so on..

ehild

 

[Femme_physics]

Ask: "Can we use voltage dividers when current flows to the base of a transistor from the middle?"

If bigger currents could flow, you should not use voltage dividers.
If no significant current flows (for instance to an op-amp) you can use voltage dividers.

But should I only apply the question of Voltage Divider to transistor? It doesn't seem to be accurate in any case.

It looks right! Substitute IC=0.8 mA and VBE=0.6 V, cancel I2 from the first two equations and so on..

ehild

Here,

Made it to 3 equations and used my calculator function to solve it:

http://img835.imageshack.us/img835/8786/solutionx1.jpg

http://img163.imageshack.us/img163/6042/solutionx2.jpg

If I got this right I can get the rest of it easily.

 

[I like Serena]

But should I only apply the question of Voltage Divider to transistor? It doesn't seem to be accurate in any case.

Transistor is the only ambivalent case I can think of.

In other cases it's usually obvious that the current is either not negligible, or there is no significant current.

 

[Femme_physics]

So, let's make it clear,

If it's visually obvious that the current is not negligible, I don't use voltage divider.

If it's visually obvious that the current is negligible, I can use voltage divider.

Yes?

 

[I like Serena]

So, let's make it clear,

If it's visually obvious that the current is not negligible, I don't use voltage divider.

If it's visually obvious that the current is negligible, I can use voltage divider.

Yes?

Yep!

("Visually obvious" :D)

 

[Femme_physics]

So, I am a little preplexed about cases where it would be visually obvious, any signs I can pick up on? Tips?

My presumption:

If Vout has really high resistance or approaching infinity-- I can use voltage divider.

If Vout has a low resistance to no resistance-- I can't use voltage divider.

 

[I like Serena]

So, I am a little preplexed about cases where it would be visually obvious, any signs I can pick up on? Tips?

My presumption:

If Vout has really high resistance or approaching infinity-- I can use voltage divider.

If Vout has a low resistance to no resistance-- I can't use voltage divider.

Those are exactly right.
I do assume that when you said that Vout has really high resistance, you meant the anything connected to Vout has a really high resistance (or is not connected at all).

Another one is if you have a connection to the input of an op-amp.

 

[Femme_physics]

I do assume that when you said that Vout has really high resistance, you meant the anything connected to Vout has a really high resistance (or is not connected at all).

Yes, but actually, I'd like to better define really high-resistance. Of course that Op-Amp is a given, as its resistance approaches infinity, but what about 200k Ohms? 300k ohms?

 

[ehild]

Here,

Made it to 3 equations and used my calculator function to solve it:

http://img835.imageshack.us/img835/8786/solutionx1.jpg

http://img163.imageshack.us/img163/6042/solutionx2.jpg

If I got this right I can get the rest of it easily.

It looks right.

ehild

 

[Femme_physics]

They are right.

Thanks

Voltage divider is good as practical technique to set the appropriate voltage. It is not a calculation method.

But in comparator op-amps?

 

[I like Serena]

Yes, but actually, I'd like to better define really high-resistance. Of course that Op-Amp is a given, as its resistance approaches infinity, but what about 200k Ohms? 300k ohms?

Suppose you have a circuit as in your earlier drawing:

http://img542.imageshack.us/img542/7336/760t.jpg

If R1 and R2 are both about 1 kΩ, and R3 is about 100 kΩ, you can neglect the current through R3.As a rule of thumb, when the current through R3 would be less than 2% of the current through both R1 and R2, you can neglect that current.

This is the case if the resistance of R3 is more than 50 times as high as both the resistance of R1 and R2.

 

[Femme_physics]

I hope you don't mind me asking but can you give me the reference to these numbers? I find this to be very important.

 

[I like Serena]

I hope you don't mind me asking but can you give me the reference to these numbers? I find this to be very important.

I just did a quick google search, but did not find any clear references.
Perhaps someone else here can say something about it.
Or else you can start a new thread about it.
I'm pretty sure such a thread will get plenty of responses.

As I said, I know that as a rule of thumb 2% is considered negligible, and sometimes 5%, depending on the field of science.
I can only applaud your request for a reference - it's the scientific way!

 

[Femme_physics]

Thanks ILS :) I did ask on elec engineering forum.

Going back to the question here. Is there anyway I can verify the solution, or is there really only one way to solve it?

 

[I like Serena]

Going back to the question here. Is there anyway I can verify the solution, or is there really only one way to solve it?

Well, assuming you can use the voltage divider, you already found Vbb.
Now what would Ib be, considering the path it would take to the emitter of the transistor?

 

[Femme_physics]

Wait wait, but if i do that first off id get a "different" result from what i do with KLV and KCL.. So what is the point of doing it?

And didnt we agree that voltage divider is not valid here?



PS i can only reply tomorrow going to sleep.. Thanks everyone !

 

[I like Serena]

Wait wait, but if i do that first off id get a "different" result from what i do with KLV and KCL.. So what is the point of doing it?

The point is that it is a verification.
Your result should be close to the answer you got with KVL and KCL.

And didnt we agree that voltage divider is not valid here?

Weren't you going to ask your teacher?

Voltage divider is usable, since the current to the base of a transistor is usually low.
A typical approach is to assume it is low enough, do the numbers, and in retrospect see if it is low enough.
(And if it's not, we can still try and be creative. You're pretty creative aren't you?)

PS i can only reply tomorrow going to sleep.. Thanks everyone !

Sleep well then! :zzz:

 

[Femme_physics]

The point is that it is a verification.
Your result should be close to the answer you got with KVL and KCL

But if I use voltage divider I thought it means Ib = 0,

which consequently means my result is indeterminate when I calculate beta since I can't divide by zero!

Weren't you going to ask your teacher?

Voltage divider is usable, since the current to the base of a transistor is usually low.
A typical approach is to assume it is low enough, do the numbers, and in retrospect see if it is low enough.
(And if it's not, we can still try and be creative. You're pretty creative aren't you?)

I will ask my teacher today at any rate, but based on what I just said earlier I thought that voltage divider makes Ib = 0 and therefor it makes the results no make sense.

 

[ehild]

It is easy to derive the output voltage of a voltage divider in terms of the "leaking current" Ib.

V_{out}=V_{in}\frac{R_2}{R_1+R_2}+I_b\frac{R_1R_2}{R_1+R_2}

The first term is the output of an unloaded voltage divider. The second term is the "error", ΔV. The relative error is
\frac{\Delta V}{V_{ideal}}=\frac{I_b}{V_{in}}R_1

A few percent error is acceptable in case of such transistor circuits. If you want to make it less than 1%, and Ib=0.02 mA, Vin = 16 V, R1<8 kΩ. Choose R1=8 kΩ. Ib=0.02 mA, IE=0.82 mA, Vout should be 1.42V. It follows that R2=0.77 kΩ.
In real transistor circuits, the resistors of the voltage divider are of kΩ-s.

ehild

 

[I like Serena]

But if I use voltage divider I thought it means Ib = 0,

which consequently means my result is indeterminate when I calculate beta since I can't divide by zero!

I will ask my teacher today at any rate, but based on what I just said earlier I thought that voltage divider makes Ib = 0 and therefor it makes the results no make sense.

Not quite.
Using voltage divider implies you assume Ib≈0 and is much smaller than I1 and I2.

From there you can calculate Vbb.
Once you have Vbb you can calculate Ib.

Anyway, I consider this particular circuit a lost cause.
Ib is certainly not negligible, which you can already see from the KVL/KCL results that you got.

 

[Femme_physics]

Ib≈0

Approximately zero, but not really?

From there you can calculate Vbb.
Once you have Vbb you can calculate Ib.

You mean I can first calculate Ir2 and then eventually calculate Ib.

Ib is certainly not negligible, which you can already see from the KVL/KCL results that you got.

Ok, fine, I did, and I got 0.0202 mA. A result approved by ehild.

HOWEVER, doing it via Voltage Divider yields a result far higher to Ib than the result using KVL and KCL!

http://img404.imageshack.us/img404/990/ibagain.jpg

0.886 mA!

That's more like Ib≈1 if anything!

 

[I like Serena]

Approximately zero, but not really?

Yep.
We "say" that Ib≈0 if it is much smaller than I2.

Ok, fine, I did, and I got 0.0202 mA. A result approved by ehild.

Aha! You're using references!
To ehild in this case.

HOWEVER, doing it via Voltage Divider yields a result far higher to Ib than the result using KVL and KCL!

http://img404.imageshack.us/img404/990/ibagain.jpg

0.886 mA!

That's more like Ib≈1 if anything!

Good! You did the numbers!

With voltage divider you found that Ib=0.886 mA and you also found I2=0.0202 mA.
So you can see that Ib is not much smaller than I2.
It is much bigger!

You found the valuable result that in this case you can't use voltage divider.

 

[Femme_physics]

Yep.
We "say" that Ib≈0 if it is much smaller than I2.

I don't recall you saying that "if it is much smaller than I2" part before.

Aha! You're using references!
To ehild in this case.

hehe yep

Good! You did the numbers!

With voltage divider you found that Ib=0.886 mA and you also found I2=0.0202 mA.
So you can see that Ib is not much smaller than I2.
It is much bigger!

You found the valuable result that in this case you can't use voltage divider.

So if I2 is bigger than Ib we can't use voltage divider?

 

[I like Serena]

I don't recall you saying that "if it is much smaller than I2" part before.

I didn't?
Let's see...

As a rule of thumb, when the current through R3 would be less than 2% of the current through both R1 and R2, you can neglect that current.

This is the case if the resistance of R3 is more than 50 times as high as both the resistance of R1 and R2.

Yes I did!
I have a reference.

So if I2 is bigger than Ib we can't use voltage divider?

It's the other way around.
More specifically, if Ib is bigger than 2% of I2 we can't use voltage divider.

 

[Femme_physics]

Yes I did!
I have a reference.

Damn, that's playing a theme tonight Alright, I'm busted.

It's the other way around.
More specifically, if Ib is bigger than 2% of I2 we can't use voltage divider.

And we're not just talking about Ib, yes? We're talking about the "leakage current" of Vout, except in our case it's Ib.

 

[NascentOxygen]

But if I use voltage divider I thought it means Ib = 0,

In practical terms, it means that I_B is << I₂, and so as far as the voltage divider is concerned, I_B has negligible effect on its voltages and currents. But that says nothing about the transistor's operation; it certainly doesn't say that the transistor's I_B = 0.0000

I've been puzzled by the way you have been using the term "voltage divider" so looked back through earlier posts to this thread. I can see now what you mean by it. The conventional meaning of "voltage divider" is the series arrangement of two resistors, simply that. You have been using it to mean the numerical fraction R1·R2/(R1+R2). So you'll understand why I was astonished by you questioning whether use of the voltage divider was even valid here, or anywhere for that matter. I was thinking "of course using two resistors is a good design technique", whereas you were thinking something quite different.

I will ask my teacher today at any rate, but based on what I just said earlier I thought that voltage divider makes Ib = 0 and therefor it makes the results no make sense.

I see you are deep in conversation on this very thing. I hope the misunderstanding has been cleared up.

There really aren't that many design situations where one can rely on a potential divider being subject to negligible loading, and inputs to op-amps and FETs are about the only cases that come to mind.

 

[Femme_physics]

I've been puzzled by the way you have been using the term "voltage divider" so looked back through earlier posts to this thread. I can see now what you mean by it. The conventional meaning of "voltage divider" is the series arrangement of two resistors, simply that. You have been using it to mean the numerical fraction R1·R2/(R1+R2). So you'll understand why I was astonished by you questioning whether use of the voltage divider was even valid here, or anywhere for that matter. I was thinking "of course using two resistors is a good design technique", whereas you were thinking something quite different.

hehe. I see. Well, I'm trying to solve a problem not design a circuit...yet! :)

I went by the way wiki demonstrated voltage divider to be, mainly focused on that formula I presented. I supposed though "Resistive divider" is a better term. I'm really not sure how to call it.

There really aren't that many design situations where one can rely on a potential divider being subject to negligible loading, and inputs to op-amps and FETs are about the only cases that come to mind.

I don't know what FETS are but certainly Op-Amps comparators are the only case I recall myself using voltage divider without anyone calling foul.

I see you are deep in conversation on this very thing. I hope the misunderstanding has been cleared up.

Yes it has. Big thanks to ILS , you, ehild and everyone else

 

[I like Serena]

I don't know what FETS are but certainly Op-Amps comparators are the only case I recall myself using voltage divider without anyone calling foul.

A FET is a transistor! :p

 

[Femme_physics]

We only learned about one type of transistors! But thanks :)