Collector current of an NPN common emitter circuit

[MrAl]

MrAl, Hello again



There are, of course, people who "have done it". People have done a lot of things. But - the question is if it makes sense or not. And I doubt, if the results "are really "usable".
Do you consider the temperature effects which even can destroy the transistor? I never have seen an amplifier stage in common emitter configuration without any dc feedback, did you?
If you don`t believe me - here are some statements to be found in the "electronic bible" (H&H, The Art of Electronics):

* "The grounded emitter amplifier should be avoided - eccept in circuits with overall negative feedback";
* "The grounded emitter amplifier is difficult to bias. It might be tempting just to apply a voltage that gives the right quiescent current. That won´t work because of the temperature dependence of Vbe.
* "Such unstable biasing is useless."


Sorry - I don`t understand what you mean because you always speak about "it". What is better than what?



I think, now we come to the "core" of the problem - and I am not sure if you aren`t mixing up two different things: (1) The physical truth and (2) the technical method.

1.) At first, there is the question, if the current Ic of a BJT is PHYSICALLY controlled by the voltage Vbe or by the base current Ib. Only one single answer is possible because the transistor physics cannot depend on external circuitry. I know, that two different explanations can be found in textbooks - however, only one can be correct. I think, the people who are "advocates" of voltage control (Ratch and myself) have some good arguments, see for example the link Ratch has given elsewhere (I think, a very useful link, in particular post#10):

https://cr4.globalspec.com/thread/68055/voltage-vs-current

If you are interested, I can give you a reference from Berkeley University explaining why the BJT is physically voltage controlled.
But - this question has nothing to do with the METHOD how the transistor is biased.

2.) There are two different methods a BJT can be biased: Voltage driven or current driven (not "controlled"). What means "current driven"? We use a large resistor Rb and a relatively large dc voltage to realize a voltage divider between Rb and the dc resistance Rbe of the B-E path. In our engineering community it is a common agreement to call this method "current driven", in spite of the fact that it is not really a current source.
The current Ib is primarily determined by the resistor Rb - that`s all. And, of course, there is a useful relationship Ic=B*Ib. But this formula does not mean that Ib controls Ic. It`s just a fixed relation.
However, the current Ic still is controlled by the voltage, which is developed across the B-E path.

Hello again,


Jony has explained what i am looking for here. If you like you can include a link to the Berkley info and i'll read it, but our view depends on what we are looking at so we also need to distinguish the internal circuit view from the external circuit view.

You said that we cant calculate the grounded emitter circuit because there is too much variation, although i would accept a variation.

But then i posted anther circuit in post #19 that has an emitter resistor, yet we can still see that Vbe is a factor and so we still have the same situation right? The only difference is the external circuit view, where we see a better circuit. But surely we still have to figure out a way to bias this circuit correct?

 

[audioguru]

The extremely crude circuit does not have an emitter resistor and a voltage divider for the base so its output current is seriously affected by the large range of hFE, the wide range of Vbe and the temperature.

If you want the circuit to work properly with any or a combination of these variables then add an emitter resistor and a voltage divider for the base.

 

[Winterstone]

Jony has explained what i am looking for here. If you like you can include a link to the Berkley info and i'll read it, but our view depends on what we are looking at so we also need to distinguish the internal circuit view from the external circuit view.

You said that we cant calculate the grounded emitter circuit because there is too much variation, although i would accept a variation.

But then i posted anther circuit in post #19 that has an emitter resistor, yet we can still see that Vbe is a factor and so we still have the same situation right? The only difference is the external circuit view, where we see a better circuit. But surely we still have to figure out a way to bias this circuit correct?

Hello MrAl,

Is the above quote the only comment you can give to my long explanations - in particular to the last two points? Disappointing!

Regarding your last sentence:
I already gave my comment (see post#21) related to the circuit as given in post#19.
For my opinion it makes absolutely no sense to discuss a circuit that nobody will use (se post#21) because - and this is common knowledge as can be found in relevant textbooks ! - it cannot compete with other known bias principles.

Quote Audioguru: If you want the circuit to work properly with any or a combination of these variables then add an emitter resistor and a voltage divider for the base.

Thank you Audioguru for this comment. That`s what I am saying since 3 days (in this and another related thread).

W.

 

[audioguru]

Since the discussed circuit is garbage then maybe we should change the base resistor position to provide some negative feedback from the collector.
But then the voltage gain is dependent on the source impedance.

 

[MrAl]

Hello again,


So neither of you two guys can analyze the circuit unless it is exactly of the topology the way you want it be, apparently because the variables are not as well behaved as you would like variables to be. That's very interesting. That must mean that you also do not know how to test a transistor for Beta, because after all it's not fixed the way you need it to be so you could be able to analyze it all nice and comfortable like

Would you guys like it if i added another resistor so there are two resistors (as a divider) on the base as well as one on the emitter? But then what else will you ask for after that?

 

[ericgibbs]

Since the discussed circuit is garbage then maybe we should change the base resistor position to provide some negative feedback from the collector.
But then the voltage gain is dependent on the source impedance.

hi agu,

Being garbage is the whole point of the posted circuit.

We are trying to highlight the common sense practical approach in determining the required biassing current for a well behaved circuit.

I suspect like you, I consult the transistors datasheet for the hFE quoted value at the collector current I require, add an emitter resistor to fix the stage 'Q' point gain.

Add an emitter resistor bypass capacitor to give increased gain over the desired bandwidth.

Eric

 

[Winterstone]

So neither of you two guys can analyze the circuit unless it is exactly of the topology the way you want it be, apparently because the variables are not as well behaved as you would like variables to be. That's very interesting.

MrAl,

do you really think, I cannot analyze the given circuit? Don`t you believe that I am able to apply Ohm`s law for Rb*Ib and to add to voltages (Vbe+Ie*Re) ?
What is the background of your question? Are we in a Kindergarten?
As a matter of fact, I am not motivated to answer such a question.
I do not want you to add a resistor elsewhere. That`s your job. If you or somebody else has a question that makes sense - I try to answer.
I have documented my view (including calculation of resistor values) for proper BJT biasing several times in this and the foregoing thread.
So - what do you want? To prove if I have heard about Ohm`s law?

W.

 

[Winterstone]

We are trying to highlight the common sense practical approach in determining the required biassing current for a well behaved circuit.
I suspect like you, I consult the transistors datasheet for the hFE quoted value at the collector current I require, add an emitter resistor to fix the stage 'Q' point gain.

Eric, there is only one practical approach if you like to bias the BJT with a current.
(I am aware that this is "engineering language", because in fact we still bias the B-E path with a voltage. As I have explained in a previous contribution, we realize a voltage divider consisting of Rb and the DC resistance of the B-E path).

As mentioned already by Audioguru, the classical (and the only reasonable) approach is to provide current feedback by connecting the base resistor Rb at the collector node (or to divide the collector resistor in two parts and use middle node).

It is very simple to list the three possible stabilization methods:
* Current biasing with Rb - together with voltage-controlled current feedback (connect Rb to the collector),
* Voltage biasing with a base divider - together with current-controlled voltage feedback (emitter resistor Re)
* Voltage biasing with a base divider - together with voltage-controlled voltage feedback (connect the resistive divider to the collector)

These are basics from control theory because you cannot add (compare) a voltage with a current and vice versa at the input of an amplifier (remember the classical feedback model)

And what is your approach: "Add an emitter resistor to fix the stage 'Q' point gain".

That means: Current biasing with Rb - together with current-controlled voltage feedback.
This approach does not stabilize the Q point - as explained, for example, in "The Art of Electronics" ("bad biasing"); didn`t you read my reply#21 ?

Winterstone

 

[audioguru]

If I am going to make only ONE circuit then I might simply test the hFE of a bunch of transistors and select ONE. Or test the hFE of only ONE transistor then design an extremely simple circuit that will work properly only with a transistor with an hFE of that ONE. Its temperature must be held within a narrow range.

Instead, why not design the transistor circuit with good biasing so that the variables have very limited effect then thousands or millions of circuits can be built and they all work properly without testing the hFE of all the transistors and selecting only a few.

 

[ericgibbs]

If I am going to make only ONE circuit then I might simply test the hFE of a bunch of transistors and select ONE. Or test the hFE of only ONE transistor then design an extremely simple circuit that will work properly only with a transistor with an hFE of that ONE. Its temperature must be held within a narrow range.

Instead, why not design the transistor circuit with good biasing so that the variables have very limited effect then thousands or millions of circuits can be built and they all work properly without testing the hFE of all the transistors and selecting only a few.

hi agu,

I agree 100%

Whats your preferred method of measuring the hFE of the transistor batch, which gives the best indication of hFE.?

Eric

 

[audioguru]

hi agu,

I agree 100%

Whats your preferred method of measuring the hFE of the transistor batch, which gives the best indication of hFE.?

Occasionally, I used a cheap multimeter to measure the hFE of transistors.

It probably used the same extremely simple circuit we are talking about:
1) if the transistor is saturated then it has a high hFE and-or a low Vbe and-or it is warm.
2) if the transistor is cutoff then it has a low hFE and-or a high Vbe and-or it is cool.

 

[MrAl]

Hi,


Eric:
Lets see if they like the new circuit that they had been asking for

BTW Winterstone you said you did the circuit i posted with the emitter resistor in your post #21 but i didnt see that there. Perhaps you did not like that circuit, so here is a new one more to your liking perhaps.

I think that maybe audioguru and Winterstone will see the light here soon. I agree that proper biasing is a good idea for a practical circuit so that we can hopefully make the circuit work without having to hand pick the transistor for example. But the questions here are a little different as you will see. In the mean time, i have drawn up a new circuit that seems to satisfy audioguru and Winterstone in their unending quest for the better biased circuit. Now you both have one, lets see what you can do with it

Attached is the circuit you have been waiting for. It has an added emitter resistor and also now has two reasonable valued resistors on the base to bias it. The values are as follows:
Rc=1000 ohms
Re=100 ohms
Rb1=54970 ohms
Rb2=7000 ohms
and also:
Vbe=0.65 volts as before.
Vs=10v as before too.

So there you go, the circuit you've always wanted. Now calculate Ic. Or do you not like that circuit either?

 

[Claude Abraham]

Sorry to be so late replying to this op question. What is Ic was the question on post #1. With a 0.65V CVS (constant voltage source) directly connected to the b-e junction, the Ic value will be indeterminate for a short time, then the bjt will likely incinerate.

Placing a CVS right to the b-e junction incurs thermal runaway. Most practitioners are familiar with Ebers-Moll equations. If the b-e junction driver is constant voltage, then the Ic value based on the forced value of Vbe is simply given by E-M as :

Ic = alpha_n*Ies*(exp(Vbe/Vt)-1) - Ics*(exp(Vbc/Vt)-1).

This equation was published in 1954 and has been in use since. Gummel-Poon eventually replaced E-M by adding a factor for Early voltage, but for now we can use the above equation E-M.

For some reason this equation is usually chopped up into pieces. The 2nd term is almost always omitted, but if the bjt is saturated, the 2nd term is all-important. The b-c junction is also relevant since a bjt is really 2 bjt devices in 1. The "collector" (lighter doped n region for an npn) is the collector in the right side up bjt. The collector is also the emitter for the upside down bjt, where the "emitter" (heavier doped n region) is the upside down device collector).

I bring this up because w/o knowing base current Ib, we don't know if the npn device is saturated, or active. If in the active region, then the E-M 2nd term is small enough to neglect, so the 1st term suffices as

Ic = alpha_n*Ies*(exp(Vbe/Vt)-1). Here, "alpha_n" is the Ic/Ie value in normal (right side up) operation, and upside down device can be neglected. Herein alpha_n can be called just "alpha". So if active region is assumed, the 1 term equation holds. But the "Ies" factor in front, is the saturation current value of the b-e p-n junction, which varies w/ specimen and w/ temp as well.

The variation of Ies w/ temp alone, is far greater than part to part and temp variations due to beta (h[sub]FE[/sub]). In addition to variation, Ies has a positive temperature coefficient, herein called "tempco". As temp increases, Ies increases non-linearly to a large extent. So if a CVS is connected directly to b-e junction, we get Ie and Ic per Ebers-Moll. Since current through the device raises its temp, the junction temp increases due to power dissipated.

Ies also increases due to temp increase, which greatly raises Ic. At higher temp, Vt does increase, making the exp(Vbe/Vt) smaller, but not enough to offset the increase in Ies. Thus bjt junction temp increases, Ies increases, Ic increases, power dissipation increases, temp increases, Ies increases, Ic increases, well you get the rest.

We have thermal runaway. This is why constant voltage is never directly connected to a bjt b-e junction. Thermal instability and runaway are inevitable. The circuit in post #1 can never be employed, regardless of application, i.e. signal gain, switching, log amp, etc.

Later in in this thread, some posters illustrated resistive divider types of biasing schemes, that should prove beneficial. The key to designing a bjt stage whose behavior is predictable and thermally stable is to control Ie, the emitter current. Ic = alpha*Ie is the transistor action equation. Alpha is around 0.99 +/- 0.01, so bjt behavior is very stable when alpha is employed as the control parameter.

Using Ie to control Ic is the way pros do it. Using Ib to control Ic is not good, because of beta variation. Ic = beta*Ib is thermally stable, i.e. a CCS (constant current source) forcing a base current value results in an Ic value dependent on beta, which has specimen and temp variation.

Using Vbe to control Ic is never done for any industrial application. I believe this question is the basis of a more theoretical discussion as the circuit in post #1 is not usable at all. I will elaborate if needed. Cheers.

 

[audioguru]

An antique 2n2222A transistor can have a very wide range of hFE from 75 to 325 then I think the emitter resistor value should be 1/10th the collector resistor value, not 1/100th.
A more modern 2N3904 has a much smaller range of hFE from 100 to 300 and can use a small emitter resistor value.

 

[Winterstone]

So there you go, the circuit you've always wanted. Now calculate Ic. Or do you not like that circuit either?

Hello,

MrAl, may I direct your attention to post#16 of this thread, where I have analyzed/calculated exactly this circuit you want me to calculate now?
Do you remember that it is not the first time that I have asked you: Why don`t you respond to my postings?
Perhaps you haven`t read them?
May I add, that I really don`t understand your ironic attitude ?
From a technical point of view it is not too important which circuit I like or dislike. Technical properties (performance, exactness, stability) are the points that matter only.

But I am very happy that - finally - you have arrived at the biasing scheme that I have promoted several times in this thread (see again post#16).

With kind regards
W.

 

[Winterstone]

Ies also increases due to temp increase, which greatly raises Ic. At higher temp, Vt does increase, making the exp(Vbe/Vt) smaller, but not enough to offset the increase in Ies. Thus bjt junction temp increases, Ies increases, Ic increases, power dissipation increases, temp increases, Ies increases, Ic increases, well you get the rest.

We have thermal runaway. This is why constant voltage is never directly connected to a bjt b-e junction. Thermal instability and runaway are inevitable. The circuit in post #1 can never be employed, regardless of application, i.e. signal gain, switching, log amp, etc.

Later in in this thread, some posters illustrated resistive divider types of biasing schemes, that should prove beneficial. The key to designing a bjt stage whose behavior is predictable and thermally stable is to control Ie, the emitter current. Ic = alpha*Ie is the transistor action equation. Alpha is around 0.99 +/- 0.01, so bjt behavior is very stable when alpha is employed as the control parameter.

Using Ie to control Ic is the way pros do it. Using Ib to control Ic is not good, because of beta variation. Ic = beta*Ib is thermally stable, i.e. a CCS (constant current source) forcing a base current value results in an Ic value dependent on beta, which has specimen and temp variation.

Using Vbe to control Ic is never done for any industrial application. I believe this question is the basis of a more theoretical discussion as the circuit in post #1 is not usable at all. I will elaborate if needed. Cheers.

To C. Abraham:

I must confess, that I didn`t read the first parts of your rather long contribution (because it contains basic considerations) - however I fully agree to the conclusions/summary as quoted above.
I hope this summary, hopefully, will help to bring this - in some parts not fruitful (but polemic) - discussion to an end.
Thank you and regards

W.l

 

[MrAl]

An antique 2n2222A transistor can have a very wide range of hFE from 75 to 325 then I think the emitter resistor value should be 1/10th the collector resistor value, not 1/100th.
A more modern 2N3904 has a much smaller range of hFE from 100 to 300 and can use a small emitter resistor value.

I am not sure what you are talking about here, is this for the new circuit? I had already stated that the values are as follows:
Rc=1000 ohms
Re=100 ohms
Rb1=54970 ohms
Rb2=7000 ohms
and also:
Vbe=0.65 volts as before.
Vs=10v as before too.

As predicted you now no longer like the emitter resistor even though it matches your specs, and now you attack the type of transistor too. I dont understand that at all.


Winterstone:
I see no analysis of this new circuit ANYWHERE in this thread. And in post #16 you apparently have done a simulation not an analysis, and that is a different circuit because it has different values.

 

[Winterstone]

Winterstone:
I see no analysis of this new circuit ANYWHERE in this thread. And in post #16 you apparently have done a simulation not an analysis, and that is a different circuit because it has different values.

Perhaps it would help to read my post #16 a second time.
I did NOT "apparently" have done a simulation - I rather have calculated (and this word even appears in post#16) the circuit.
For your understanding I will summarize what I have done:
* The classical design starts with selection of a suitable value for Ic (in my example I have chosen Ic=1mA) and a corresponding amount of feedback (example: 1 V DC feedback with Re=1kOhm)
* Based on these values: Calculation of a resistive divider for the two cases: Vbe=0.65 and Vbe=0,7 Volts.
* Finally, checking this design using simulation.
* Result: The calculated resistor values allow Ic=0.92 mA (Vbe=0.65V) resp. Ic==0.96 mA (Vbe=0.7V)
* Summary: The classical method to bias a BJT (resistive divider with dc feedback) allows a design that is very stable against parameter tolerances and other unwanted deviations and uncertainties.

That was the goal of this exercise.
What do you want me to do in addition? To use "different values" (as proposed by you) ? I am not your pupil.

 

[audioguru]

I am not sure what you are talking about here, is this for the new circuit? I had already stated that the values are as follows:
Rc=1000 ohms
Re=100 ohms
Rb1=54970 ohms
Rb2=7000 ohms
and also:
Vbe=0.65 volts as before.
Vs=10v as before too.

As predicted you now no longer like the emitter resistor even though it matches your specs, and now you attack the type of transistor too. I dont understand that at all.

Sorry. I looked at your schematic yesterday but it had no resistor values and I mistakenly used 10k for the collector resistor instead of 1k.
But why not use a modern 2N3904 that has a smaller range of hFE instead of the antique 2N2222A (in an old metal case??) that has a wide range of hFE?
Also, the modern 2N3904 datasheet has many more spec's and graphs than the antique transistor.

 

[MrAl]

Hello Winterstone and audioguru,

Perhaps it would help to read my post #16 a second time.
I did NOT "apparently" have done a simulation - I rather have calculated (and this word even appears in post#16) the circuit.
For your understanding I will summarize what I have done:
* The classical design starts with selection of a suitable value for Ic (in my example I have chosen Ic=1mA) and a corresponding amount of feedback (example: 1 V DC feedback with Re=1kOhm)
* Based on these values: Calculation of a resistive divider for the two cases: Vbe=0.65 and Vbe=0,7 Volts.
* Finally, checking this design using simulation.
* Result: The calculated resistor values allow Ic=0.92 mA (Vbe=0.65V) resp. Ic==0.96 mA (Vbe=0.7V)
* Summary: The classical method to bias a BJT (resistive divider with dc feedback) allows a design that is very stable against parameter tolerances and other unwanted deviations and uncertainties.

That was the goal of this exercise.
What do you want me to do in addition? To use "different values" (as proposed by you) ? I am not your pupil.

Winterstone:
Yes you are, now get to work or else you get graded with an "F" for this semester and i have to call your parents on the phone
Seriously though, you didnt want to do the original circuit and asked for a new circuit, and i provided one and you did not like that one either, so i provided another one with all the things you wanted and now you say you dont want to do it. Why even bother responding in this thread if you dont want to do the circuit posted in the thread? Start your own thread and do the circuit you want done then. It's that simple.

Audioguru,
So you dont want to do the circuit as is yet again? Ok then i'll look into the other transistor you suggested and maybe incorporate that next.

All:
See, as predicted there is still no analysis of the circuit i wanted to see (except by those who already had done so and didnt constantly complain about the circuit)