Transistor Biasing: Is the Base Current Constant in Both Biasing Schemes?

[Bassalisk]

Consider this picture, a) and b).
[PLAIN]http://pokit.org/get/f0af323a103c1d29539836143a7784a1.jpg

Now I am getting confused a bit here.

a) I know that the first picture represents a bias for current source. And we change the value on the potentiometer the current will stay the same. (considering we won't go too far and put transistor into saturation)

From the computations I got, and simulations in national instruments, I can say that the base current is adjusting accordingly in order to keep the current constant. I can handle that idea, I think.

b) Will the same happen here? If I change the potentiometer. Consider that resistor connected to base to be thevenin equivalent of those 2 voltage dividers and that voltage source at base to be also be within thevins theorem. And every other parameter stays the same.

But I have the feeling that the base current here is determined only by that voltage source at base and that resistor, hence it won't change when I change the potentiometer, hence the current in collector must fall.

Are those to schemes one and the same? Will I have the same effect?

 

[yungman]

As in the diagram, you have a resistor at the emitter to 0V. That and the base voltage is really what determine the collector current. b) is the thev. eq. of a), they both set up the voltage on the base.

The reason we always use a resistor at the emitter because the current gain of the transistor is not stable, the emitter resistor really serve as the negative feedback. Say the voltage at the base give the base current, which in turn create the emitter current. as emitter current increase the voltage across the emitter resistor increase. At the point where Vbe is about 0.7V, any further increase in emitter will cause the Vbe to drop and decrease the emitter current. It is this negative feedback that keep the current constant at the collector. This kind of circuit eliminate the dependency of base current and become more a voltage dependent.

I don't think I explain this very well, look up a basic book of bipolar transistor and they explain much better than me the role of emitter degenerating resistor and how it take away the variable of the current gain of a transistor.

 

[Bassalisk]

As in the diagram, you have a resistor at the emitter to 0V. That and the base voltage is really what determine the collector current. b) is the thev. eq. of a), they both set up the voltage on the base.

The reason we always use a resistor at the emitter because the current gain of the transistor is not stable, the emitter resistor really serve as the negative feedback. Say the voltage at the base give the base current, which in turn create the emitter current. as emitter current increase the voltage across the emitter resistor increase. At the point where Vbe is about 0.7V, any further increase in emitter will cause the Vbe to drop and decrease the emitter current. It is this negative feedback that keep the current constant at the collector. This kind of circuit eliminate the dependency of base current and become more a voltage dependent.

I don't think I explain this very well, look up a basic book of bipolar transistor and they explain much better than me the role of emitter degenerating resistor and how it take away the variable of the current gain of a transistor.

But can you confirm that the base current has to rise, in order to keep the current in collector constant, whilst changing the resistance of collector resistor high up?

 

[es1]

But can you confirm that the base current has to rise, in order to keep the current in collector constant, whilst changing the resistance of collector resistor high up?

Base current will be the same for constant collector current.
Ib = Ic / Bf.

Now technically the voltage at the collector will change and the transistor will experience a change in Bf due to this which will in turn alter Ib (this is known as the early effect). But I think you can probably ignore this non-ideality if this is for homework and if you were not yet aware of it.

http://en.wikipedia.org/wiki/Early_effect

 

[Bassalisk]

Base current will be the same for constant collector current.
Ib = Ic / Bf.

Now technically the voltage at the collector will change and the transistor will experience a change in Bf due to this which will in turn alter Ib (this is known as the early effect). But I think you can probably ignore this non-ideality if this is for homework and if you were not yet aware of it.

http://en.wikipedia.org/wiki/Early_effect

Nah this is my trying to understand things. Homework was long done. Now look, If i change the collector resistance, and the current through collector stays the SAME, something HAS TO GIVE. I feel like the base current will increase, in order to keep the collector current constant for all resistances. Assume that you can change the collector resistance so much, that you don't get transistor into saturation.

 

[yungman]

But can you confirm that the base current has to rise, in order to keep the current in collector constant, whilst changing the resistance of collector resistor high up?

If you look at the data sheet of the specific transistor, you can find a graph that show given a particular biasing condition ( like your circuit), how the collector current change with change in Vce. You'll find the change is very very little and in the first approx, you take it as constant.

Of cause it really vary very little and yes you need to change the base bias ever so little to keep the collector constant. But really nobody worry about this with BJT. When you talk about FETs, then it's a different story, because change of drain current vs Vds is a lot more prominent and cannot be ignored.

Unless you are getting into semi conductor physics, just assume the collector current is constant. Take our words, I design bipolar ICs and all different transistor circuits, never once I have to worry about this.

 

[es1]

In real transistors something does give. If Rc gets larger (without saturating the transistor) and the current stays the same then the voltage drop on the resistor goes up and the voltage on the collector goes down. The CB depletion width changes in response to the changing Vc. This changes the gain of the transistor which in turn changes the base current. This is known as the early effect and was linked to above.

However, the transistor is designed to have high gain and high immunity to this non-ideality. Say for example Ic=10mA and Bf dropped from 200 to 150 due to decreasing Vc. Then Ib changes from 50uA to 66uA. If the biasing resistors are small enough then this change in current can be ignored.

This is why there is the design rule "make divider current 10x the base current" in the app below, for example.
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/npncc.html#c3

 

[yungman]

Yes, in semi conductor physics, it's the early voltage and it is very big for BJT...not so big for FETs.

For the op, if you look at the collector curve, the slope on the straight portion is the output resistance. And if you extrapolate the straight portion ( the slope is the output resistance) of different collector curves( different collector currents), they all come to a same point and the voltage is the early voltage. For NPN, the early voltage Va is always negative.

 

[jim hardy]

Mr Bassalisk ,
i think sometimes in teaching we forget to stress the very basics

contemplate for a moment that resistor between the emitter and ckt common.
it makes your circuit into a closed loop with negative feedback.


current through the base-emitter junction follows that formula i = e^(qv/kt) ,
where i is current, e is 2.718, q and k are constants, v is base-emitter voltage, and t is temperature...
now look where voltage is in that term - it's up in the exponent - so a miniscule change in that voltage really affects base current.

Sooo -- if collector current changes ever so slightly so does emitter current
--- which changes voltage at emitter --
---which changes voltage across emitter-base junction which substantially changes base current,,,

and that base current change is in direction to push collector current back toward where it was... which is negative feedback. (negative feeedback opposes change)
In circuit design that's the purpose of emitter resistor, to set collector current by feedback through base.
It reduces gain because that's what negative feedback does

but you have correctly picked up that something is missing from everyday explanations of this.

It's not that a transistor inherently has constant collector current
but we can use its current gain to make it do so by adding an emitter resistor and setting voltage across that resistor and base junction...
realizing that the base junction's share of that voltage will be small ,
and that's how we bias a transistor when we use it in a circuit..

in ac amplifiers you'll see emitter resistor bypassed by a capacitor which removes the negative feedback for AC , allowing higher gain.

Have you studied feeedback yet? It's a fascinating world. Take a course in automatic controls.

i hope I'm not beating a comatose horse here. If enough explanations are tossed out here one of them will click with your internal thinking processes and ah AHA! moment will result. I hope this one helps.

 

[Bassalisk]

THANK YOU GUYS! I just got it. This slight change of current in base is due to early effect right? That slight incline in BJT characteristic ? But in IDEAL BJT, current would be constant through base, until it reaches saturation.

Transistor will reach saturation when Uce is below certain value. And that certain value is determined by voltage drop across Ropt. When it takes too much voltage drop, transistor gets bottomed(saturated) and THEN base is not the function of beta anymore.

Did I understand this?

 

[Bassalisk]

Mr Bassalisk ,
i think sometimes in teaching we forget to stress the very basics

contemplate for a moment that resistor between the emitter and ckt common.
it makes your circuit into a closed loop with negative feedback.current through the base-emitter junction follows that formula i = e^(qv/kt) ,
where i is current, e is 2.718, q and k are constants, v is base-emitter voltage, and t is temperature...
now look where voltage is in that term - it's up in the exponent - so a miniscule change in that voltage really affects base current.

Sooo -- if collector current changes ever so slightly so does emitter current
--- which changes voltage at emitter --
---which changes voltage across emitter-base junction which substantially changes base current,,,

and that base current change is in direction to push collector current back toward where it was... which is negative feedback. (negative feeedback opposes change)
In circuit design that's the purpose of emitter resistor, to set collector current by feedback through base.
It reduces gain because that's what negative feedback does

but you have correctly picked up that something is missing from everyday explanations of this.

It's not that a transistor inherently has constant collector current
but we can use its current gain to make it do so by adding an emitter resistor and setting voltage across that resistor and base junction...
realizing that the base junction's share of that voltage will be small ,
and that's how we bias a transistor when we use it in a circuit..

in ac amplifiers you'll see emitter resistor bypassed by a capacitor which removes the negative feedback for AC , allowing higher gain.

Have you studied feeedback yet? It's a fascinating world. Take a course in automatic controls.

i hope I'm not beating a comatose horse here. If enough explanations are tossed out here one of them will click with your internal thinking processes and ah AHA! moment will result. I hope this one helps.

We just had a lesson about feedbacks. I didn't get the point of it as much. I might as well go study feedback a little more, since I see you guys are talking a lot about it. :)

 

[jim hardy]

i guess we were typing at same time...

i don't know what you mean by 'early effect'

but sounds like you're getting it!

we're pullin for you -

old jim

 

[Bassalisk]

i guess we were typing at same time...

i don't know what you mean by 'early effect'

but sounds like you're getting it!

we're pullin for you -

old jim

Thank you! got to go now. I have another lab. With oscilloscopes!

 

[Bassalisk]

Oh the sick irony! I went to my third lab today, exercise about oscilloscopes and amplifiers. When wouldn't you know it... power went down. We couldn't do that lab, because we didn't have electricity.
Apparently power company was working on some transformer or something.

I laughed so hard, because at electrical engineering faculty, we didn't have electricity :D

 

[yungman]

:rofl:

 

[yungman]

THANK YOU GUYS! I just got it. This slight change of current in base is due to early effect right? That slight incline in BJT characteristic ? But in IDEAL BJT, current would be constant through base, until it reaches saturation.

Transistor will reach saturation when Uce is below certain value. And that certain value is determined by voltage drop across Ropt. When it takes too much voltage drop, transistor gets bottomed(saturated) and THEN base is not the function of beta anymore.

Did I understand this?

No, I just read this! Slight change of COLLECTOR current due to early voltage effect, not the base current. I thought you ask whether you need to change the base voltage to keep collector current constant when you vary the collector resistor.

 

[Bassalisk]

No, I just read this! Slight change of COLLECTOR current due to early voltage effect, not the base current. I thought you ask whether you need to change the base voltage to keep collector current constant when you vary the collector resistor.

Yea that was a long shot with the Early effect :D. But yes that was the question:
"I thought you ask whether you need to change the base voltage to keep collector current constant when you vary the collector resistor."

And I understood, that in IDEAL transistor you don't have to. But jim told me that it has something to do with feedback, which is why the current in base is changing slightly in real transistor.

You pointed out this too, just I don't understand fully the concept of feedback.

I just realized that everything I knew about transistors is slowly drowning. I have a feeling that I will have to invest a lot of energy to fully understand this small yet complicated component.

 

[yungman]

I tried to explain the feedback in post#2 but I don't think I did a good job and don't know how to explain it better. Remember it is the voltage between the base and the emitter that govern the collector current, not just the base voltage. The emitter resistor serve as the feedback to stabilize the transistor so it does not drift that much. Hope someone else can write it better to explain to you.

BTW, nobody claim this 3 legged animal is easy! Take your time, keep asking question here and people will try to answer. One book that I studied this subject is by Malvino. It is the easiest and the best book on transistor I've seen.

 

[Bassalisk]

I tried to explain the feedback in post#2 but I don't think I did a good job and don't know how to explain it better. Remember it is the voltage between the base and the emitter that govern the collector current, not just the base voltage. The emitter resistor serve as the feedback to stabilize the transistor so it does not drift that much. Hope someone else can write it better to explain to you.

BTW, nobody claim this 3 legged animal is easy! Take your time, keep asking question here and people will try to answer. One book that I studied this subject is by Malvino. It is the easiest and the best book on transistor I've seen.

You really made me laugh with this 3 legged animal hahahahah.

Don't worry. I truly appreciate yours, and any other expert help that I can get here. Because, to be honest, today there are few people that are willing to even try to explain something to you.

People like You, Jim and countless others, make the students life more easy, where new concepts are not that naive.

I will definitely check out that book, thank you for recommendation.

 

[es1]

No, I just read this! Slight change of COLLECTOR current due to early voltage effect, not the base current. I thought you ask whether you need to change the base voltage to keep collector current constant when you vary the collector resistor.

Slight change in 'forward gain' due to early voltage. This changes both the collector and the base current technically. If we assume the feedback mechanism hold Ie constant and the biasing network was correctly picked to hold Vb constant.

Ie is actually the controlled variable here. As the emitter and collector current are very close in a forward active high gain NPN usually we just consider them to be the same.

This link does a good job of running through the math.
http://www.stanford.edu/class/ee122/Handouts/EE113_Course_Notes_Rev0.pdf
Start at section 6.

Google for "common emitter feedback" and you'll get a ton more links which explain it in various ways.

The above calculations assume an ideal NPN and therefore B is constant but it is pretty easy to see from their math what would happen if B was reduced due to early effect.

So I think the answer to the question you were really trying to ask, what breaks as I increase the pot's resistance, is B is reduced. Or, the NPN becomes a poorer amplifier even though it is still forward active.

 

[Bassalisk]

Slight change in 'forward gain' due to early voltage. This changes both the collector and the base current technically.

The feedback mechanism in this circuit is designed to try to keep the emitter current constant. As the emitter and collector current are very close in a forward active high gain NPN usually we just consider them to be the same.

This link does a good job of running through the math.
http://www.allaboutcircuits.com/vol_3/chpt_4/10.html

Google for "common emitter feedback" and you'll get a ton more links which explain it in various ways.

The above calculations assume an ideal NPN and therefore B is constant but it is pretty easy to see from their math what would happen if B was reduced due to early effect.

So I think the answer to the question you were really trying to ask, what breaks as I increase the pot's resistance, is B is reduced. Or, the NPN becomes a poorer amplifier even though it is still forward active.

So my hunch about Early effect was actually right? wow :D

 

[es1]

So my hunch about Early effect was actually right? wow :D

Yup. I thought your original understanding of the NPN operation was right.

The feedback concept in this circuit is something else.

I edited my post a bit to try to keep the wording clearer. There are a number of moving pieces and as one adds details things get murkier so I wanted to make sure to state my assumptions. Hopefully the newer version, and the new link, is better.

 

[Bassalisk]

Yup. I thought your original understanding of the NPN operation was right.

The feedback concept in this circuit is something else.

I edited my post a bit to try to keep the wording clearer. There are a number of moving pieces and as one adds details things get murkier so I wanted to make sure to state my assumptions. Hopefully the newer version, and the new link, is better.

Already reading it. Many thanks.

 

[yungman]

That little 3 legged animal can be as complicated as you want. The place where I find hard to explain is exactly what ES1 said, the collector current change due to the collector voltage as you change the collector resistor with the base voltage constant must mean there must be some change to the base current and the gain of the transistor. If you take into consideration of all these, it will never end. That is the reason I mentioned about the Malvino book. It present a very simplistic view of the transistor...and surprisingly it is good enough to design very advanced circuits.

You really chase to the bottom of this, you'll find it very long and unnecessary hard. From the question you ask, I think the Melvino book is perfect for you. Not only that, this book get the best introduction of op-amp I ever seen including Bode Plot.

Baby steps, learn the basic of transistors and then go back and dig into it. It can fill up a whole thick just on this little animal. You are not going to get any simple answer on this.

Read the book, ask questions here. If I have a chance, I'll try to look for the book on Amazon and see whether you can take a peek inside the section on DC biasing and AC equivalent of NPN.

 

[Bassalisk]

That little 3 legged animal can be as complicated as you want. The place where I find hard to explain is exactly what ES1 said, the collector current change due to the collector voltage as you change the collector resistor with the base voltage constant must mean there must be some change to the base current and the gain of the transistor. If you take into consideration of all these, it will never end. That is the reason I mentioned about the Malvino book. It present a very simplistic view of the transistor...and surprisingly it is good enough to design very advanced circuits.

You really chase to the bottom of this, you'll find it very long and unnecessary hard. From the question you ask, I think the Melvino book is perfect for you. Not only that, this book get the best introduction of op-amp I ever seen including Bode Plot.

Baby steps, learn the basic of transistors and then go back and dig into it. It can fill up a whole thick just on this little animal. You are not going to get any simple answer on this.

Read the book, ask questions here. If I have a chance, I'll try to look for the book on Amazon and see whether you can take a peek inside the section on DC biasing and AC equivalent of NPN.

Can you give me the title of the book?

 

[yungman]

I was trying to see whether Amazon will have "Look inside this book" option on this book, sadly they don't. Here is what I think is the book, mine is a very old edition 30 years back, it look nothing like this. I just remember the name of the author. Try this one, it is only about $4 used, so it would not be too much money if this is not the one. In fact I just bought one for myself.

https://www.amazon.com/dp/0028028341/?tag=pfamazon01-20

Look for the DC bias and AC equivalent section. It give you a lot of approximation to make it easy to learn. It does not even get into the early voltage which you don't really need. I am currently designing some electronics used in guitar and I am using the knowledge gained in that book. Just hope it's the right one.

 

[Bassalisk]

I was trying to see whether Amazon will have "Look inside this book" option on this book, sadly they don't. Here is what I think is the book, mine is a very old edition 30 years back, it look nothing like this. I just remember the name of the author. Try this one, it is only about $4 used, so it would not be too much money if this is not the one. In fact I just bought one for myself.

https://www.amazon.com/dp/0028028341/?tag=pfamazon01-20

Look for the DC bias and AC equivalent section. It give you a lot of approximation to make it easy to learn. It does not even get into the early voltage which you don't really need. I am currently designing some electronics used in guitar and I am using the knowledge gained in that book. Just hope it's the right one.

I just LOVE good books. Might sound geeky but between google and a good book, I always go for a good book.

Thank you for the research, I will surely buy it.

 

[yungman]

I just LOVE good books. Might sound geeky but between google and a good book, I always go for a good book.

Thank you for the research, I will surely buy it.

I hope you have not order that yet, I received the book, it is the exercise book for the textbook, not the textbook! No wonder it is so cheap!

 

[jim hardy]

i'd recommend the ancient Texas Instruments book "Transistor Circuit Design"

a google showed it in a lot of used bookstores.

While on the subject their wonderful "OpAmps for Everyone" is available as a download
from TI's library

search for slod006b.pdf and add it to your library...

 

[Bassalisk]

No I didn't order from there, because I don't think Amazon can send to my country. I found it on ebay.

http://www.ebay.co.uk/itm/Electronic-Principles-Albert-Paul-Malvino-/180738521739?pt=Non_Fiction&hash=item2a14dafa8b

 

[Bassalisk]

i'd recommend the ancient Texas Instruments book "Transistor Circuit Design"

a google showed it in a lot of used bookstores.

While on the subject their wonderful "OpAmps for Everyone" is available as a download
from TI's library

search for slod006b.pdf and add it to your library...

Already did. Thank you. I am learning waaaay more than I need to pass my exam. This course I am taking is very narrow, although it should be since I am studying telecommunications. But I cannot just accept things like, negative feedback, differential amplifiers and so on. Need to have a nice explanation for it :)

 

[yungman]

No I didn't order from there, because I don't think Amazon can send to my country. I found it on ebay.

http://www.ebay.co.uk/itm/Electronic-Principles-Albert-Paul-Malvino-/180738521739?pt=Non_Fiction&hash=item2a14dafa8b

It is the correct author, make sure you get the textbook, not the experiments book that I bought.:grumpy:

 

[Bassalisk]

Btw. the book, Electronic Principles by Malvino, is just out of this world. I just wish I had it when I started learning transistors. Its making my life just 10 times more easy.

Thank you very much.

 

[yungman]

I am glad you like it. Make sure you study the opamp and Bode Plot. It is very very good. It is so good that I even tried to buy one and get the wrong one.

Don't think this is too simple to be useful, I designed heavy duty transistors circuits and integrated circuits for years, the stuff works. that little r'e and Vt that is 25ohm for 1mA and 25mV and room temperature is all you need to know and to scale the r'e with current.