Transistors circuit - Basic questions

[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!

 

[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?