Engineering Transistors circuit - Basic questions

[Femme_physics]

You can say that VB ≅ R2/(R1+R2)⋅VCC if you wish,
but on building the circuit and measuring VB you will find VB is not what you thought, so other parameters will likewise be a little different from intended. VB is not going to be what you hoped precisely because base current is not zero.

Do I necessarily presume that the voltage in Ib is zero when I make this voltage divider? Are you sure?

This effect of loading a resistive divider is nothing peculiar to a transistor; anything that
draws current from (or into) the junction of two resistors will change the voltage there.

Then what's the point of voltage divider? It's confusing to me how this formula even works in general cases! Because, there is practically always coming coming out of the junction!

 

[NascentOxygen]

Do I necessarily presume that the voltage in Ib is zero when I make this voltage divider?

By saying VB ≅ R2/(R1+R2)⋅VCC
you are implicitly assuming I_B is zero (or at least is so small in comparison with the "wasted current" that drains through the divider to ground, viz., VCC/(R1+R2), as to be safely ignored).

That unloaded resistive divider formula is derived on the assumption that all the current that passes through R1 also passes through R2. You can prove this for yourself.

Or, if you wish to view it a different way, you can stick to that formula but you must modify the value of R2 so you include in R2 a parallel resistance equivalent to the base input resistance. (So there is just no escaping it, unfortunately.)

Then what's the point of voltage divider?

It provides a cheap way to drop the voltage from VCC down to something lower. And it's stable and predictable. And it gives a more stable Q-point than does a single large value resistor from VCC directly to the base.

It's confusing to me how this formula even works in general cases! Because, there is practically always coming coming out of the junction!

So that must always be allowed for in the design calculations.

Sorry to be the bearer of bad news, Femme_physics.

 

[Femme_physics]

By saying VB ≅ R2/(R1+R2)⋅VCC
you are implicitly assuming IB is zero (or at least is so small in comparison with VCC/(R1+R2) as to be safely ignored).

That unloaded resistive divider formula is derived on the assumption that all the current that passes through R1 also passes through R2.

That's news to me.

So pretty much anytime I use voltage divider I can ignore the current in the junction. Such as, in this case, I ignore R3:

http://img210.imageshack.us/img210/4652/49303121.jpg Now, in the pic below here, in the upper case, I CANNOT use voltage divider because of a simple logic that overrides the voltage divider-- current will always prefer to flow where there is no resistance.

In the lower case, current will not flow via R3.

http://img710.imageshack.us/img710/5858/r1r2r3.jpg

So, really, for the lower case I can just do this:
Therefor if I know the value of the resistors I just figure out I1. Hence, I don't really see the point or usefulness of a voltage divider.

Or, if you wish to view it a different way, you can stick to that formula but you must modify the value of R2 so you include in R2 a parallel resistance equivalent to the base input resistance. (So there is just no escaping it, unfortunately.)

I'm not sure what you mean by that.

So that must always be allowed for in the design calculations.

I'm not sure what it means, "must always be allowed for in design calculations"?

It provides a cheap way to drop the voltage from VCC down to something lower. And it's stable and predictable. And it gives a more stable Q-point than does a single large value resistor from VCC directly to the base.

But it's INACCURATE!Stable and predicable is one thing, but it's not realistic, doesn't the whole thing goes down the drain?

And VCC doesn't change as far as I know. How does it drop it down?

Sorry, too confused. Many of my ideas here shuttered in an instant! so, trying to see the logic here.

 

[NascentOxygen]

So pretty much anytime I use voltage divider I can ignore the current in the junction. Such as, in this case, I ignore R3:

Apparently you have been ignoring it in the past, but this is reckless. You really cannot ignore the load connected to the junction of the resistors.

Now, in the pic below here, in the upper case, I CANNOT use voltage divider because of a simple logic that overrides the voltage divider-- current will always prefer to flow where there is no resistance.

R3 has a short circuit across it. There cannot be a potential difference across a short circuit, so there is likewise no potential difference across R3. With no potential across R3 there can be no current through it. By the laws of physics!

In the lower case, current will not flow via R3.

Why do you say that? If there is a potential difference across it, then current can/will/must flow though it according to Ohm's Law.

Hence, I don't really see the point or usefulness of a voltage divider.

Resistive dividers are incredibly useful. They allow us to obtain any voltage we like less than VCC just by wasting a bit of current draining away through R1+R2. The alternative would be to have lots of batteries, each of a different voltage right where we needed it! (Not really, as that's not possible nor practical either.)

But it's INACCURATE!Stable and predicable is one thing, but it's not realistic, doesn't the whole thing goes down the drain?

It's only inaccurate if you make unrealistic assumptions in your design. Why assume there is no current into the base when we clearly recognize there is?!

And VCC doesn't change as far as I know. How does it drop it down?

VCC doesn't change, no. But by using different resistive dividers you can obtain any voltage you desire, so long as it's < VCC. That sounds enormously useful to me!

Sorry, too confused. Many of my ideas here shuttered in an instant! so, trying to see the logic here.

What time is it where you are? I get the feeling that you've been studying hard and it must be getting late. I think you will see things in a more positive light in the morning. You are doing well. In a clearer light I think you will be surprised that you didn't notice the loading effect on a potential divider before now. There is a lot to learn, some will inevitably get overlooked and need to be re-visited.

Onwards towards that 3000th post!

 

[Femme_physics]

I just made a huge post, made a wrong move--hit my laptop, and the entire thing crashed :(

SIGH

Frustrating. I wish forums had auto-draft-save system like Gmail.

Anyway, I'll start rewriting...

Onwards towards that 3000th post!

Haven't been studying hard enough, apparently!

It's actually morning here! I'm in my work at the coast 1st-aid booth but as long as no one breaks a leg or goes into cardiac arrest (Goodness forbid) I'm here. You know, "I Like Serena" even stopped for a visit here once! But I won't go into details ;)

Apparently you have been ignoring it in the past, but this is reckless. You really cannot ignore the load connected to the junction of the resistors.

So Voltage Divider you might say is irrelevant in the first case I presented.

R3 has a short circuit across it. There cannot be a potential difference across a short circuit, so there is likewise no potential difference across R3. With no potential across R3 there can be no current through it. By the laws of physics!

True, Voltage Divider is unusable here as well (in the shortcircuited example where R3 doesn't exist) since the entire voltage of the circuit falls on R1.

Why do you say that? If there is a potential difference across it, then current can/will/must flow though it according to Ohm's Law.

So, even in this is a case where I can't use a voltage divider as well. If I use a voltage divider, I assume that I1 passes through R2 as well ->

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

Whereas in reality and in accordance to KVL we know that current splits in the junction and that R2 gets I2. So what's written in the pic is false.I really then fail to see at what circumtances can I use voltage divider.

Resistive dividers are incredibly useful. They allow us to obtain any voltage we like less than VCC just by wasting a bit of current draining away through R1+R2. The alternative would be to have lots of batteries, each of a different voltage right where we needed it! (Not really, as that's not possible nor practical either.)

Fine, they can redefine the value of VCC (i.e. Vin), but only if Vbb (Vout) is already defined. If Vbb is not defined, it becomes kinda useless again?

It's only inaccurate if you make unrealistic assumptions in your design. Why assume there is no current into the base when we clearly recognize there is?!

I understand this point.

VCC doesn't change, no. But by using different resistive dividers you can obtain any voltage you desire, so long as it's < VCC. That sounds enormously useful to me!

So we use the ratio of resistors to define Vout?

 

[I like Serena]

Do you mean that I can't officially use Voltage Divider?

That depends on what you're being taught.
In engineering results usually don't have to be perfect.

Ahh...I thought Vbb is always responsible for Ib. I see that's not the case. I'll fix that

Well, Vbb is responsible for Ib, but you need to use the path through the transistor to the emittor, which is how the current flows.

but I don't know if my use of voltage divider is valid now

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

 

[I like Serena]

You know, "I Like Serena" even stopped for a visit here once! But I won't go into details ;)

Yeah, that was fun!
To finally see someone I've been interacting with intensively.

 

[ehild]

Forget voltage divider for a moment and remember good old KCL and KVL. How are I_R1, I_R2 and IB relared at the node B?
There is also the transistor equation: IE=IC+IB, you have written in the first sheet.
And you have found IC already IC=0.8 mA..
Write up the potential U_B. You can do it in three ways. Once it is U_B=U_CC-300(kΩ)I_R1. And it is also U_B=50(kΩ)I_R2. And here comes the best of all: U_B=U_BE +1(kΩ)I_E.

You have five unknowns and five equations, but all are very simple. You are an expert multi-variable-equation-solver already , you can manage it.

ehild

 

[NascentOxygen]

So we use the ratio of resistors to define Vout?

I'm sure you have met voltage dividers in many guises, but perhaps not always by that name.

Let's revise their design.

We have a string of two resistors, R1 and R2, hung between VCC and ground, respectively. We agree that the open circuit voltage at the junction of the resistors is VCC*(R2/(R1+R2).

This little arrangement can be represented as a Thevenin equivalent comprising an ideal voltage source of VCC*(R2/(R1+R2) volts in series with an impedance of R1//R2 ohms. (If you work it out, that equates to R1·R2/(R1+R2) ohms.)

The voltage VCC*(R2/(R1+R2) you are happy with; the existence of the impedance seems to have taken you aback. That's the internal resistance of the voltage source, and it accounts for the fact that if current is drawn from (or injected into) that voltage divider, its voltage will inevitably change.

Voltage dividers are a compromise, like almost everything in engineering. https://www.physicsforums.com/images/icons/icon6.gif You CAN give it a low internal resistance by choosing for R1 and R2 to be low value resistors, but the big drawback here is that now you are wasting a lot of current straight through R1 & R2 to ground. (And who wants to be replacing the batteries in their radio/camera/phone/clock every few hours because the batteries are going flat too quickly?)

 

[Femme_physics]

That depends on what you're being taught.
In engineering results usually don't have to be perfect.

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"?

Well, Vbb is responsible for Ib, but you need to use the path through the transistor to the emittor, which is how the current flows.

I see. :)

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

Yep, hence here :)

Yeah, that was fun!
To finally see someone I've been interacting with intensively.

*grins* :) I agree, that was great!

Forget voltage divider for a moment and remember good old KCL and KVL. How are I_R1, I_R2 and IB relared at the node B?
There is also the transistor equation: IE=IC+IB, you have written in the first sheet.
And you have found IC already IC=0.8 mA..
Write up the potential U_B. You can do it in three ways. Once it is U_B=U_CC-300(kΩ)I_R1. And it is also U_B=50(kΩ)I_R2. And here comes the best of all: U_B=U_BE +1(kΩ)I_E.

You have five unknowns and five equations, but all are very simple. You are an expert multi-variable-equation-solver already , you can manage it.

ehild

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

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

Right?

I'm sure you have met voltage dividers in many guises, but perhaps not always by that name.

Let's revise their design.

We have a string of two resistors, R1 and R2, hung between VCC and ground, respectively. We agree that the open circuit voltage at the junction of the resistors is VCC*(R2/(R1+R2).

This little arrangement can be represented as a Thevenin equivalent comprising an ideal voltage source of VCC*(R2/(R1+R2) volts in series with an impedance of R1//R2 ohms. (If you work it out, that equates to R1·R2/(R1+R2) ohms.)

The voltage VCC*(R2/(R1+R2) you are happy with; the existence of the impedance seems to have taken you aback. That's the internal resistance of the voltage source, and it accounts for the fact that if current is drawn from (or injected into) that voltage divider, its voltage will inevitably change.

Voltage dividers are a compromise, like almost everything in engineering. You CAN give it a low internal resistance by choosing for R1 and R2 to be low value resistors, but the big drawback here is that now you are wasting a lot of current straight through R1 & R2 to ground. (And who wants to be replacing the batteries in their radio/camera/phone/clock every few hours because the batteries are going flat too quickly?)

Thanks for the explanation :) Got it.

We didn't learn about about Thevenin equivalent though btw. Not in our course material.

Voltage dividers seem to be a lame "compromise" compared to KVL and KCL, which is immaculate and always correct provided we take everything int account, right?

 

[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 :)

 

[ehild]

A FET is a transistor! :p

FETis Field Effect Transistor, with high input resistance.
The one in the problem is bipolar transistor, and it has low input resistance.
A voltage divider is two resistors in series connected to a voltage source and using the voltage between one end of one resistor and the common point of the pair of resistors. A potentiometer can serve as voltage divider.

ehild

 

[NascentOxygen]

A potentiometer can serve as voltage divider.

Indeed it can. A potentiometer can serve as a variable voltage divider.

Potentiometer...not to be confused with a voltmeter...and definitely not to be confused with a voltameter. :shy:

 

[ehild]

Indeed it can. A potentiometer can serve as a variable voltage divider.

Potentiometer...not to be confused with a voltmeter...

which works on the principle of voltage divider.

and definitely not to be confused with a voltameter. :shy:

which is definitely not a voltage divider.

ehild

 

[Femme_physics]

So I talked to my teacher. He told me that I'm wrong and he doesn't know what I'm talking about and that voltage divider would give me the same result for Ib as Kirchhoff laws. I showed him all my papers trying to prove him wrong but he said I'm not using voltage divider correctly, that I should use the formula

R1 x R2 / R1 + R2

To create a new resistor, RBB, that Ib goes through, and then I can just use this resistor while ignoring R1 and R2. The result that turned up is indeed the same result as Ib with Kirchhoff's...

So I don't see how anything you said make sense if we simply didn't use the right formula.

 

[NascentOxygen]

So I talked to my teacher. He told me that I'm wrong and he doesn't know what I'm talking about

What a nasty fellow!

and that voltage divider would give me the same result for Ib as Kirchhoff laws.

That's because, as I explained, you have been using the phrase "voltage divider" to mean something completely different from the usual meaning. (I think we needn't go over all that again.)

I showed him all my papers trying to prove him wrong but he said I'm not using voltage divider correctly, that I should use the formula

R1 x R2 / R1 + R2

To create a new resistor, RBB, that Ib goes through, and then I can just use this resistor while ignoring R1 and R2.

That's what I showed as the Thevenin equivalent, but because you haven't studied that yet, there is no necessity to analyze the circuit that way. Whatever way the circuit is analyzed, if it is valid and is done correctly it will arrive at the same answer. It is all basic mathematics; there is no room for magic or skullduggery.

The result that turned up is indeed the same result as Ib with Kirchhoff's...

Well, it couldn't be called the Thevenin equivalent if it didn't give equivalent results.

So I don't see how anything you said make sense if we simply didn't use the right formula.

I can't see myself in this picture; I'm sure this must be a reference to some other "you".

So, is it all sorted now? Or (I hesitate to ask) still some confusion? :shy:

 

[Femme_physics]

What a nasty fellow!

Well, he said that HE doesn't know what I'm talking about not that I don't know what I'm talking about. He's ok :)

I can't see myself in this picture; I'm sure this must be a reference to some other "you".

So, is it all sorted now? Or (I hesitate to ask) still some confusion?

LOL. All I can do is thank you, you keep me engaged, asking, and curious,and with a supply of intelligent answers. What we did is not called voltage divider then, it's called "Thévenin's theorem", that's the only difference?

 

[NascentOxygen]

LOL. All I can do is thank you, you keep me engaged, asking, and curious,and with a supply of intelligent answers.

Sure don't see myself here, either. Really must be some other "you" this time.

What we did is not called voltage divider then, it's called "Thévenin's theorem", that's the only difference?

Back in this post I stated that it is possible to view the two resistors between Vcc and ground as equivalent to a voltage source in series with a single resistor. When applied to transistor biasing, as here, that Thevenin voltage source is referred to as V_BB. But as someone else had introduced a weird way to name voltage from base to emitter as V_BB I didn't compound the confusion by pointing out that's not right and the symbol V_BB should refer to the Thevenin base voltage (in precisely the same way that V_CC truly does refer to the Thevenin voltage of the collector supply).

I think you should give this topic a rest, and put your energies towards other topics. Come back to this in a few weeks, and with a fresh start you'll wonder how you managed to make such heavy work of it first time around.

You know enough now to be able to sort it out yourself, just pencil, paper, calculator and Ohms Law, and a quiet location away from distractions.