Engineering Transistors circuit - Basic questions

[Femme_physics]

Homework Statement

Given the following transistoric circuit. The transistor acts in the active region. In addition: VBB = 0.6V

http://img855.imageshack.us/img855/9901/newcircuit.jpg

The Attempt at a Solution

Before I get there I have two questions:

1) What does it mean that the transistor acts in the active region?
2) VBB is not the drawing. Why does it matter to me what it equals to if I can't see it?

 

[gneill]

When a transistor is "acting in the active region", it means that it is operating in the region of its characteristic curves where it is behaving essentially linearly. That is, it is not biased to cutoff (no collector current flowing) or to saturation (where small changes in base current are ineffective at altering the collector-emitter current).

VBB is the designation for the base voltage with respect to circuit ground.

 

[Femme_physics]

When a transistor is "acting in the active region", it means that it is operating in the region of its characteristic curves where it is behaving essentially linearly. That is, it is not biased to cutoff (no collector current flowing) or to saturation (where small changes in base current are ineffective at altering the collector-emitter current).

Hmm...I'm not sure what you mean by "not biased to cutoff"...in fact that explanation kinda confused me. Can you try to write it differently maybe it would help, please?

VBB is the designation for the base voltage with respect to circuit ground.

Hmmm.

I've circled in red where I think VBB is, VEE is and VCC is. Is that correct?

http://img269.imageshack.us/img269/8585/cccircledinredok.jpg

 

[gneill]

Biased to cutoff means that the base voltage drops below the threshold required for base current to flow; the base-emitter junction behaves like a diode, and you need to forward bias a diode in order for current to flow. If the base-emitter voltage (Vbe) drops below this forward bias threshold (about 0.7V for a silicon transistor), then base current will stop and and the collector-emitter current will be shut off, too. The transistor then "looks like" an open circuit for collector-emitter current, and the base can no longer actively control that current.

Similar things happen at what is called "saturation". So much base current flows that the transistor is "fully on" -- the tap's open as much as it can go! No further increase in base current can cause more collector-emitter current to flow, and that current becomes essentially insensitive to small changes in base current.

Regarding the V_CC, V_BB, and V_EE designations, have a look here:

They designate the DC voltage with respect to ground at the collector, base, and emitter terminals of the transistor. It is perhaps an unfortunate fact that traditionally the supply voltage that provides the supply voltage for the circuit has also come to be called "Vcc"... so one needs to be a bit careful to distinguish between that supply voltage and the V_CC that exits at the transistor terminal when a collector load resistor stands in the path between them. Usually the distinction is made clear by statement or context in a given problem.

 

[Femme_physics]

Maybe I have problem with my English here, but "forward bias a diode"? don't understand what it could mean.

Bias means "prejudice" or "influence opinions"...are we trying to pretend electrical components have feelings?

If the base-emitter voltage (Vbe) drops below this forward bias threshold (about 0.7V for a silicon transistor), then base current will stop and and the collector-emitter current will be shut off, too. The transistor then "looks like" an open circuit for collector-emitter current, and the base can no longer actively control that current.

We're actually treating it as an ideal transistor, so I think it means that the 0.7V doesn't exist?

In fact, just to clear things up, is my view of where the 0.7V is correct?

http://img825.imageshack.us/img825/7752/17888088.jpg

Regarding the V_CC, V_BB, and V_EE designations, have a look here:

They designate the DC voltage with respect to ground at the collector, base, and emitter terminals of the transistor.

Ah this clears it thank you. I'll try to work on the exercise soon^^

 

[Femme_physics]

Just so I make sure, because it's sometimes hard for me to see the voltage source: is this it?

http://img651.imageshack.us/img651/9761/thecirrrrrrrrrrcuit.jpg

And I just simply use Kirchhoff laws and the transistors formulas (with the beta/IE/IC/IB stuff...) to solve it?

 

[gneill]

Maybe I have problem with my English here, but "forward bias a diode"? don't understand what it could mean.

Bias means "prejudice" or "influence opinions"...are we trying to pretend electrical components have feelings?

That's more or less it In electronics a bias is the potential difference that exists across some component, usually one designed on purpose by some arrangement of resistors (and sometimes other components) in concert with the circuit's power supply. This is referred to as the "bias network".

You recall that a diode only conducts when its anode terminal is at a greater potential than its cathode, right? When the anode is positive with respect to the cathode the diode is said to be forward biased and will allow current to pass through. On the other hand, when the anode is negative with respect to the cathode no current will flow and the diode is said to be reverse biased.

We're actually treating it as an ideal transistor, so I think it means that the 0.7V doesn't exist?

Most ideal transistor models that I've come across keep the (approximately) 0.7V junction drop in them. This is because the base current can be very sensitive to very small changes in the base voltage, and the collector current variations are β times that. You might want to refer to your course materials to see what they are suggesting in this section.

In fact, just to clear things up, is my view of where the 0.7V is correct?

http://img825.imageshack.us/img825/7752/17888088.jpg

Yes, that's it for the 0.7V. But keep in mind that it's a voltage drop that's internal to the transistor (it is in fact the voltage drop of a forward biased junction inside the transistor). In practice, when the transistor is operating in the active region this voltage varies a bit from 0.7V depending upon the magnitude of the base current. More sophisticated transistor models will take this into account. An external bias can make this voltage less until the base-emitter is no longer forward biased, shutting off the base current and hence the collector-emitter current.

You should make a point of indicating the emitter lead in your diagrams by placing the arrow on it per the conventional symbol. Note that the direction of the arrow also indicates whether the transistor "sandwich" is comprised of layers arranged NPN or PNP, which makes a difference in the required biasing arrangements for the part. The arrow indicates the direction that current will flow when the base-emitter is forward biased. Your original diagram indicated an NPN transistor, so current will pass from the base to the emitter when it's forward biased, and collector-emitter current direction is from the collector to the emitter.

Just so I make sure, because it's sometimes hard for me to see the voltage source: is this it?

http://img651.imageshack.us/img651/9761/thecirrrrrrrrrrcuit.jpg

Your original image showed the voltage source Vcc connected above the collector load resistor R3 (at the little circular connection point at the top of the circuit). So you can think of having a 16V battery (Vcc is given as 16V) connected between that point and the ground node. Again, keep in mind there's the possibility of confusing V_CC the collector potential (w.r.t. ground) and the power supply Vcc.

And I just simply use Kirchhoff laws and the transistors formulas (with the beta/IE/IC/IB stuff...) to solve it?

Yes, that's more or less it.

 

[Femme_physics]

That's more or less it In electronics a bias is the potential difference that exists across some component, usually one designed on purpose by some arrangement of resistors (and sometimes other components) in concert with the circuit's power supply. This is referred to as the "bias network"

Oh, OK. First time I hear that!

You recall that a diode only conducts when its anode terminal is at a greater potential than its cathode, right? When the anode is positive with respect to the cathode the diode is said to be forward biased and will allow current to pass through. On the other hand, when the anode is negative with respect to the cathode no current will flow and the diode is said to be reverse biased.

Thanks for the reminder, and the language tutorial

Most ideal transistor models that I've come across keep the (approximately) 0.7V junction drop in them. This is because the base current can be very sensitive to very small changes in the base voltage, and the collector current variations are β times that. You might want to refer to your course materials to see what they are suggesting in this section.

I've sent him an email yesterday, and we have a class tomorrow, so I'll find out but I believe we should ignore it.

Yes, that's it for the 0.7V. But keep in mind that it's a voltage drop that's internal to the transistor (it is in fact the voltage drop of a forward biased junction inside the transistor).

Right. So my drawing is accurate? For instance, if a current flows through the emitter to the collector or vise versa...is there a voltage drop?

You should make a point of indicating the emitter lead in your diagrams by placing the arrow on it per the conventional symbol. Note that the direction of the arrow also indicates whether the transistor "sandwich" is comprised of layers arranged NPN or PNP, which makes a difference in the required biasing arrangements for the part. The arrow indicates the direction that current will flow when the base-emitter is forward biased. Your original diagram indicated an NPN transistor, so current will pass from the base to the emitter when it's forward biased, and collector-emitter current direction is from the collector to the emitter.

You're right, I forgot to do it this time, although we were told in our course we'd only see NPN.

Your original image showed the voltage source Vcc connected above the collector load resistor R3 (at the little circular connection point at the top of the circuit).

Yes I copied it exactly.

So you can think of having a 16V battery (Vcc is given as 16V) connected between that point and the ground node.

Are you sure? Like this?

http://img52.imageshack.us/img52/4740/likethist.jpg

But why did they mark a plus and a minus like in the original image if that's not where the voltage source is?

Again, keep in mind there's the possibility of confusing VCC the collector potential (w.r.t. ground) and the power supply Vcc.

I'll keep that in mind, though I fail to understand why they'd call it in similar names.

 

[gneill]

The power supply connection (Vcc) looks like this:

Note that Vcc is not the same as V_CC.

But why did they mark a plus and a minus like in the original image if that's not where the voltage source is?

Ah! That + and - are designating where the output of the circuit is considered to be. I suppose that at some point you'll be asked to find the voltage at the collector (with respect to ground), and perhaps, with some small signal being applied at the base -- you will see that that signal has been both amplified and inverted.

I'll keep that in mind, though I fail to understand why they'd call it in similar names.

I doubt that it came about through a committee vote It's one of those things that are put down to "historical reasons".

 

[Femme_physics]

The power supply connection (Vcc) looks like this:

Note that Vcc is not the same as V_CC.

Do the two ground locations touch each other, too? Like this?

http://img42.imageshack.us/img42/8331/ummgs.jpg

I find it rather weird how you translated the original circuit to that. Let me apply your logic (or what I think is your logic) to another problem.Does this,

http://img40.imageshack.us/img40/2902/original1n.jpg

Really means that?

http://img35.imageshack.us/img35/8210/meanthatmeanthat.jpg

Ah! That + and - are designating where the output of the circuit is considered to be.

Like, if you want to connect this circuit to some other circuit or device?

I suppose that at some point you'll be asked to find the voltage at the collector (with respect to ground), and perhaps, with some small signal being applied at the base -- you will see that that signal has been both amplified and inverted.

I have the strangest feeling my mind has been both amplified and inverted just now.



Thanks.

I doubt that it came about through a committee vote It's one of those things that are put down to "historical reasons".

Oh alright. Trust old people and convention to overcomplicate things.

Anyway, thanks for now, I'm going to go to class and BBL.^^ You're great help gneill!

 

[gneill]

Do the two ground locations touch each other, too? Like this?

http://img42.imageshack.us/img42/8331/ummgs.jpg

Yes, exactly. The "ground" node is a common connection point throughout a circuit.

I find it rather weird how you translated the original circuit to that. Let me apply your logic (or what I think is your logic) to another problem.


Does this,

http://img40.imageshack.us/img40/2902/original1n.jpg

Really means that?

http://img35.imageshack.us/img35/8210/meanthatmeanthat.jpg

Yes. Note that my "circuit translation" consisted of making explicit the implied power supply, Vcc.

Like, if you want to connect this circuit to some other circuit or device?

Yes, precisely.

I have the strangest feeling my mind has been both amplified and inverted just now.



Thanks.

No problem. Always glad to help

 

[Ouabache]

Hi FP!
Are you familiar with http://en.wikipedia.org/wiki/Voltage_divider" ?
(just follow the link)
With this knowledge you can calculate
precisely what Vbb is. (Hint: it is not 0.6V)

I suspect your instructor meant to say Vbe = 0.6V (the turn on
voltage between the base & emitter junction of your transistor)

Armed with Vbb, using KVL, you could make an equation
from that point through base-emiitter junction, through Re to ground potential.
There would be only one unknown variable in that equation.
By solving, you would know the value of that variable.

 

[Femme_physics]

Yes, exactly. The "ground" node is a common connection point throughout a circuit.

But there's a difference between this:

http://img42.imageshack.us/img42/8331/ummgs.jpg

And that?

http://img683.imageshack.us/img683/4518/diff1.jpg

Right?

Which one is it?

Yes. Note that my "circuit translation" consisted of making explicit the implied power supply, Vcc.

My instructor said it's best not to use 2 Vcc's. But rather to call the arrows to the ground point you drew here

http://img840.imageshack.us/img840/900/tocall.jpg

As Vc, Vb and Ve.

Vcc, Vbb, and Vee are generally the names of voltage sources for the colllector/base/emitter respectively. That's what he told me, anyway.

Hi FP!
Are you familiar with voltage division?

I "think" I do. I may have used it in the past not knowing the name.

I suspect your instructor meant to say Vbe = 0.6V (the turn on

You're right. I talked with him about it today and there's a misprint - it's VBE = 0.6V

Armed with Vbb, using KVL, you could make an equation
from that point through base-emiitter junction, through Re to ground potential.
There would be only one unknown variable in that equation.
By solving, you would know the value of that variable.

Alright then!
http://img851.imageshack.us/img851/8580/900o.jpg

 

[gneill]

But there's a difference between this:

http://img42.imageshack.us/img42/8331/ummgs.jpg

And that?

http://img683.imageshack.us/img683/4518/diff1.jpg

Right?

Which one is it?

They are electrically identical. Any points connected by a conductor (wire) must be at the same potential, and are the same node electrically. The ground connections represent a common conduction path for anything connect to it via the "ground" symbol.

My instructor said it's best not to use 2 Vcc's. But rather to call the arrows to the ground point you drew here

http://img840.imageshack.us/img840/900/tocall.jpg

As Vc, Vb and Ve.


Vcc, Vbb, and Vee are generally the names of voltage sources for the colllector/base/emitter respectively. That's what he told me, anyway.

That works fine. There are various conventions used by different books, authors, schools, etc.. As long as you understand what particular potential or current is being referred to there should be no problem. Choose a naming scheme and stick with it throughout a given problem.

 

[gneill]

The assumed value of Vo = 8V seems a bit odd to me. Usually one wants to determine the operating point of the transistor, and hence the output voltage, from the biasing network conditions rather than the other way around. Certainly if I do so using typical assumptions about the transistor, I don't arrive at 8V for Vo.

Is there more to this problem than has been shown so far? I haven't seen an actual statement about what specifically is to be determined.

 

[I like Serena]

Alright then!
http://img851.imageshack.us/img851/8580/900o.jpg

In your second loop you write:
Sum V = 0; V_cc - R₃I_c - R_EI_E = 0

But you're forgetting the voltage V_CE here, which is not zero.

However, if you work it out properly, there's still something wrong with the numbers in this circuit. They are wrong to make physical sense, as gneill already noted.

Note that you found a beta = 0.1111.
Actually you made a mistake (as I already noted), but you should know yourself that beta is an amplification factor, typically greater than 10.

"Trust your physical cues!"

 

[Femme_physics]

Thank you. I have a test in hydraulics tuesday. I'll reply to these messages IN EXTENSIVE DETAILS on wednesday. Real thanks. You guys rock.

 

[Femme_physics]

They are electrically identical. Any points connected by a conductor (wire) must be at the same potential, and are the same node electrically. The ground connections represent a common conduction path for anything connect to it via the "ground" symbol.

You're right, I can't believe I forgot this simple fact.

That works fine. There are various conventions used by different books, authors, schools, etc.. As long as you understand what particular potential or current is being referred to there should be no problem. Choose a naming scheme and stick with it throughout a given problem.

Will do

Is there more to this problem than has been shown so far? I haven't seen an actual statement about what specifically is to be determined.

Find Beta
Find voltage on Re (VRE)
Find voltage on R2 (VR2)

you're forgetting the voltage VCE here, which is not zero.

Ah, I thought that's the link I might have been missing.

Note that you found a beta = 0.1111.
Actually you made a mistake (as I already noted), but you should know yourself that beta is an amplification factor, typically greater than 10.

Oh. 0.1111 can never be a logical result for beta?

 

[I like Serena]

Hi Fp!

Oh. 0.1111 can never be a logical result for beta?

For a regular transistor it's not a logical result.
I can't speak for all transistors though.

If you correct your calculations you'll find a reasonable result for beta.
It's such a nice round value, it has to be right!
And you'll also find reasonable results for your other values.

 

[Femme_physics]

Duly noted! Though I'm once again finding myself building too many equations and too many unknowns. Any hints?

 

[I like Serena]

Well, basically you have 2 loops.
That should give you 2 KVL and 1 KCL equation.
You've done that before.

What have you got so far?

 

[Femme_physics]

We I used 2 loops, and I used 1 KCL and looking at transistors equations as well (shouldn't I?)

http://img9.imageshack.us/img9/1741/equationsssss.jpg

 

[I like Serena]

Your first equation was right in the previous scan, now it's not.

You second equation is good now.

In your third equation you should have a plus sign instead of a multiplication dot.

I don't understand what your 4th equation is. What is I_T?
You can already find IC by looking just at R3 and the voltage difference across it.

Furthermore, you don't have R2 in any loop yet, and you should, so I guess there's a third loop you should consider.

 

[Femme_physics]

Your first equation was right in the previous scan, now it's not.

But I've added the missing link this time (Vce), how can it be wrong?

In your third equation you should have a plus sign instead of a multiplication dot.

Oops :shy:

You can already find IC by looking just at R3 and the voltage difference across it.

You're right, it must equal 8!

You second equation is good now.

Good At least that!

You can already find IC by looking just at R3 and the voltage difference across it.

You're right! I haven't thought of that! !

Furthermore, you don't have R2 in any loop yet, and you should, so I guess there's a third loop you should consider.

Duly noted. Working on it!

 

[gneill]

Say, FP, have you covered Thevenin equivalents yet? The reason I ask is that it is sometime convenient to replace the base bias network with a Thevenin equivalent voltage and series resistance.

 

[I like Serena]

But I've added the missing link this time (Vce), how can it be wrong?

Before you had 2 equations.

The first was right. This loop included Vo, which means you're not taking your loop through the transistor, and so Vce is not part of the loop.

The second equation was a loop that did go through the transistor.
That equation was missing Vce.
However, you've left this equation out now.
But that's okay, because you've included a new loop that also walks through RE.
Since you do not need to calculate Vce, you don't need the equation that has it.

You're right, it must equal 8!

Good At least that!

You're right! I haven't thought of that! !

Good!

Duly noted. Working on it!

Good, otherwise I might have to start slapping you.

 

[Femme_physics]

Say, FP, have you covered Thevenin equivalents yet? The reason I ask is that it is sometime convenient to replace the base bias network with a Thevenin equivalent voltage and series resistance.

No and I'm not sure if I should, it's not in the scope of my course. Unless it's easy. ILS, what do you think? Should I?

Before you had 2 equations.

The first was right. This loop included Vo, which means you're not taking your loop through the transistor, and so Vce is not part of the loop.

The second equation was a loop that did go through the transistor.
That equation was missing Vce.
However, you've left this equation out now.
But that's okay, because you've included a new loop that also walks through RE.
Since you do not need to calculate Vce, you don't need the equation that has it.

Oh, wait. I don't need to consider Vce because Vo already considers THAT and IRE. So my IC should still be correct. No? Just look at this.
http://img687.imageshack.us/img687/9420/thatlook.jpg

The blue dots. That's the voltage difference. It already includes Vce. So I didn't ignore it, most certainly didn't!

Good, otherwise I might have to start slapping you.

Like it dirty huh?

 

[I like Serena]

No and I'm not sure if I should, it's not in the scope of my course. Unless it's easy. ILS, what do you think? Should I?

I find them hard to apply properly, although they make the calculations easier.
For myself I prefer the calculations over breaking my head on how to apply Thevenin properly and wondering if I did it right.

Oh, wait. I don't need to consider Vce because Vo already considers THAT and IRE. So my IC should still be correct. No? Just look at this.

The blue dots. That's the voltage difference. It already includes Vce. So I didn't ignore it, most certainly didn't!

Nope. You didn't this time round.

Like it dirty huh?

Yes, I've missed dirty talking to you!

 

[Femme_physics]

haha

Nope. You didn't this time round.

So my original calculations are actually correct?

 

[I like Serena]

haha



So my original calculations are actually correct?

Nooooo!

 

[Femme_physics]

I mean this ->
http://img202.imageshack.us/img202/2527/ciranswers.jpg

What's wrong here? I was perfectly applying KVL I think.

 

[Femme_physics]

I don't understand what your 4th equation is. What is IT?

Oh, that's the total current

 

[gneill]

I find them hard to apply properly, although they make the calculations easier.
For myself I prefer the calculations over breaking my head on how to apply Thevenin properly and wondering if I did it right.

It's pretty straightforward in the case of this base bias network; only two resistors and the Vcc power supply to deal with:

The collector current, I_c is known (from Vcc, Vo, and the collector resistor R3). \beta relates this I_c to I_b, so the base circuit KVL can be expressed in terms of I_c and \beta. Solve for \beta.

 

[I like Serena]

I mean this ->

What's wrong here? I was perfectly applying KVL I think.

Yes, this one is right.
Actually, this is the one where you find IC (again! ;)).

Oh, that's the total current

Aaaah.
All right, this one won't help you much, since it introduces an extra variable IT, making the equation not so very useful.

What would be useful is another loop through R2 and RE.
With that, and the other equations you have, you should be able to solve the system.

 

[Ouabache]

It's pretty straightforward in the case of this base bias network; only two resistors and the Vcc power supply to deal with:

I'm sorry i missed the interim discussion. I was away on holiday and recently returned.
gneill's description is the standard method for analyzing this type of transistor circuit.
My earlier post suggests how to find Vb (or Vth)(the voltage applied to the base)
by http://en.wikipedia.org/wiki/Voltage_divider" (see resistive divider on the link).

However I was missing Rb (or Rth) which gneill has included (nice job!)
Rb is the combined resistance as viewed from the base of the transistor and is R1 // R2, (// = in parallel with).
You may want to take a look at the first diagram on page 6 of this http://www.kennethkuhn.com/students/ee351/bjt_bias_analysis.pdf" (posted by an EE instructor
at Univ of Alabama). He arrives at the same calculation for Vb and Rb.
This gives you sufficient information to find \beta, V(R2) and V(Re).

 

[Femme_physics]

Taking all the new comments into consideration

http://img708.imageshack.us/img708/7770/isthatright.jpg

...and I still get beta = 1/9

 

[I like Serena]

Apparently you have changed your loops yet again.
You should really make up your mind!

As it is, your loop (B) is wrong.
It contains Vbe instead of Vbc (which you don't need).

I haven't checked the rest, but they will be wrong too because of this.

 

[Femme_physics]

Corrected it to a 3 equation 3 unknowns thing. Used a calculator to solve. Is that right?

http://img831.imageshack.us/img831/4474/didididi.jpg

 

[I like Serena]

You have 4 unknowns here.
Did you mix up Ib and Ie?

So no, your results are not right.

 

[Femme_physics]

You have 4 unknowns here.
Did you mix up Ib and Ie?

So no, your results are not right.

Yes I did. Damn, a forth equation, this is insane!

 

[I like Serena]

It's not so bad...
Substitute Ie=Ib+Ic... and you're back (you already know Ic).

 

[Femme_physics]

Ok, great, because my calculator can't do 4 eq with 4 unknowns as far as I know.

So here:

http://img651.imageshack.us/img651/1443/nailedit.jpg

 

[I like Serena]

Really, you should work more careful. ;)
You made a sign mistake while substituting.
It should be: I₁ - (I_E - 0.0008) - I₂ = 0.

 

[Ouabache]

I "think" I do. I may have used it in the past not knowing the name.

Hi FP!
Are you familiar with voltage division?

Great, it is a very useful concept in circuit design.
Okay, for a little refresher...
Voltage Division

[PLAIN]http://img855.imageshack.us/img855/4493/voltagedividerresistive.jpg
(http://en.wikipedia.org/wiki/Voltage_divider#Resistive_divider")

If your Vin = Vcc and Vout is the voltage present at the node indicated,
Typically you are asked to determine Vout given Vin, R1 and R2...

Considering these components alone (with no additional circuitry),
the current flowing through R1 and R2 is the same, i1 = i2 = i

\frac {Vin}{R1+R2} = i
\frac {Vout}{R2} = i
so \frac {Vin}{R1+R2} = \frac {Vout}{R2}
rearranging terms Vout = Vin \frac {R2}{R1+R2}
(you may want to note this last one, it is the standard
expression for a two element voltage divider).
Also, as you might anticipate, there is an analog relationship
for current (current divider) but since it is not needed for your problem,
let's save that for another thread.

Lets try a practice example:
Vcc = Vin = 12Vdc
R1 = 120Kohm
R2 = 40Kohm
Vout = ?

 

[Femme_physics]

I know I'm bumping an old topic- but believe it or not I still haven't solved it, and we're going to have a big summing test on electronics this year. So, I want to fully fathom this.

First off, I realize that by employing voltage divider I shorten my path to the solution. However, I appear to be getting 2 different results for Ie...

http://img576.imageshack.us/img576/5091/given.jpg

http://img651.imageshack.us/img651/562/voltage1.jpg

http://img828.imageshack.us/img828/4785/voltage2.jpg

 

[I like Serena]

Well, a voltage divider does not work properly here, since the current Ib is leaking away.

It might still give you a reasonable approximation, since Ib will be a small current, but then you write that Ib=Vbb/R2, which is not true.
I think you're mixing it up with I2=Vbb/R2.

 

[Femme_physics]

Well, a voltage divider does not work properly here, since the current Ib is leaking away.

It might still give you a reasonable approximation, since Ib will be a small current

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

but then you write that Ib=Vbb/R2, which is not true.
I think you're mixing it up with I2=Vbb/R2.

Ahh...I thought Vbb is always responsible for Ib. I see that's not the case. I'll fix that, but I don't know if my use of voltage divider is valid now

 

[NascentOxygen]

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

An unloaded voltage divider is only a rough approximation to the actual circuit when supplying base current . The constructed circuit will not be as close to the designed parameters if you neglect the effect of I_B from the resistive divider.

I expect in an examination you would lose marks.

 

[Femme_physics]

Wait, if it was a regular circuit without Ib-- say I3, we could use it accurately in voltage divider, but because we have a transistor with its Ib, we can't? Is it because it acts in the active region?

I don't understand how come the rules of electronics I learned no longer hold up in this type of scenario.

I'm alsow worried if there is no way of avoiding the 4 equations 4 unknowns thing

 

[NascentOxygen]

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

This departs somewhat from the goal of neatly "designing" the transistors operating point.

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. The effect can be calculated, and allowed for, at the design stage--by not assuming I_B=0.

The rules of electronics always hold!