BJT amplifies current or voltage?

[Claude Abraham]

I did look it up. They are not the same thing. Ies is given by ebbers-moll, which is a very accurate model but is not being used by the OP's equation. There are more than one way to calculate IC. They are two different models using two different coefficients. Both coefficients are a functions of geometry, carrier mobility and doping levels, among other things.

Writing IE=IC/α is completely valid, and identical to your verision of IC=αIE. It's not a matter of what happens first, the equaions follow simple algebraic manipulation without losing validity.

Using a different model for analysis does not in any way attempt to invalidate ebbers-moll. It remains a relative accurate model to use, if one wishes. However, the equations that the rest of us are using gives very good results too.

Dude, you lost me. What other equations are there? There is E-M, & then the Gummel-Poon, which is a refinement of E-M which includes Early voltage influence & other properties. The relation between Ib, Vbe, Ie, Ic, alpha, & beta is well established. Also, the "Is" in the diode equation is exactly the "Ics" & "Ies" in the E-M eqns. No difference. The geometry, temp, doping, etc. determine the value.

What am I not making clear?

 

[BrownOut]

Well, there is the equation the some of us have written many times on this thread. It doesn't have a name, it's just an equation for calculating IC directly. It uses a different coefficient than the E-M model. It's used in many engineering texts, as in pretty much all of the texts I used, and evidently, the text the OP is using.

I'm not longer worried that our young scholor might be confused. From his comments, I think he understands just fine. If he has any further questions, he is welcome to PM me, and I'll explain further.

 

[Claude Abraham]

Well, there is the equation the some of us have written many times on this thread. It doesn't have a name, it's just an equation for calculating IC directly. It uses a different coefficient than the E-M model. It's used in many engineering texts, as in pretty much all of the texts I used, and evidently, the text the OP is using.

I'm not longer worried that our young scholor might be confused. From his comments, I think he understands just fine. If he has any further questions, he is welcome to PM me, and I'll explain further.

The "coefficient" is not different. In many texts, the author simply neglects the "alpha" factor in the eqn because it is close to unity, around 0.98 to 0.998. The eqn you quote is E-M. The coefficient of the exponential is "alpha*Ies". The omission of alpha is simply an acknowledgement that its value is close to 1. Same eqn, same coefficient. No brainer.

 

[MrAl]

Hi,


We might be getting off track here a little?
The main question was about why there seemed to be two views on the
transistor, and the answer was more or less that there are at least two
different approaches to the analysis of a transistor.
The way this question came up is understandable. Many devices have only
one set of equations or even just one equation to describe it. With the
transistor, he probably saw the exp() equation first and thought, "ok, that
seems reasonable, so that's the equation for the transistor", then later
starting seeing another equation, "Ic=B*Ib", and though, "hey wait a minute,
i thought the transistor equation was exp() which has voltage as input!".

It takes a little explaining to show how the equations for the transistor have
evolved over time and some of the shortcuts that came about.

 

[Claude Abraham]

Hi,


We might be getting off track here a little?
The main question was about why there seemed to be two views on the
transistor, and the answer was more or less that there are at least two
different approaches to the analysis of a transistor.
The way this question came up is understandable. Many devices have only
one set of equations or even just one equation to describe it. With the
transistor, he probably saw the exp() equation first and thought, "ok, that
seems reasonable, so that's the equation for the transistor", then later
starting seeing another equation, "Ic=B*Ib", and though, "hey wait a minute,
i thought the transistor equation was exp() which has voltage as input!".

It takes a little explaining to show how the equations for the transistor have
evolved over time and some of the shortcuts that came about.

Who is WE? I explained clearly that a bjt amplifies BOTH current & voltage, and that the gains are given by the 3 eqns I posted. I've acknowledged all 3 eqns w/ inputs of Ib, Vbe, & Ie. I stated emphatically that all 3 variables are important to bjt operation.

The conflict was over my eqn 2) which had an alpha factor, vs. another poster's eqn similar to my no. 2), but w/o the alpha. Ic is always alpha*Ie. Many texts approximate alpha as unity, which is pretty close at 0.99, and omit alpha in the eqn. 2). I simply stated that alpha is in the eqn but close to unity. To say that Ic = Ies*exp( ) is very close to correct, the error being 1 or 2%. That is my point.

No conflict at all. Neglecting alpha will produce an error of less than 2% typical. As long as the bjt has existed, that has been the case. I've read Drs. Ebers & Molls original 1954 IEEE paper several times, and little has changed except that Drs. Gummel & Poon refined the E-M model.

 

[alphacat]

Hey.
I must say I learned even more from the discussion you had here
Its good to be introduced to new equations.

In reality, what do the small signals and loads (RL's) represent?
I assume that one example for it is voice as the small signal input and speaker as the load.

 

[BrownOut]

We are off-track. I feel just a little bad about causing another contraversy LOL!

My concern was the the OP might think the eqn he was using was the same as eqn. 2 that was given and ignoring alpha. Had he reinserted alpha into his eqn, thinking that would give a better result, then went further to calculate IE as IC/alpha, he would have ended up with IE=IC, which of course would be a wrong result.

Why do I care so much about that? Because I well remember the frustration I went thru trying to learn transistors. And I remember how much a simple mistake would set me back.

 

[BrownOut]

Hey.
I must say I learned even more from the discussion you had here
Its good to be introduced to new equations.

In reality, what do the small signals and loads (RL's) represent?
I assume that one example for it is voice as the small signal input and speaker as the load.

I've never heard of a small-signal load. There are small-signal parameters such as rΠ, re, etc. If you can explain exactly what you're talking about, I might be able to help.

 

[alphacat]

Sure, I meant that the load (RL) is the one who receives that amplified small signal and uses it to provide the destination of the whole device.
Like a speaker which receives the amplified voice and turns it into sound which we can hear.

 

[BrownOut]

Sorry, I misread your statement. Yeah, voice, or an electrical representation, and a speaker are good examples of a small signal source and load. When we talk of small-signal, we only mean signals that are suffiently small that distortion from the transistors non-linear characteristics don't significantly distort the result. There are methods for analyzing what is meant by "small enough", and it's not an exact science.

 

[alphacat]

I was looking to hear some other examples besides the voice and speaker case.
It makes it much more interesting to learn about all these types of amplifiers when you treat the load not just as a resistor nor the input as a sine wave.

 

[BrownOut]

The "coefficient" is not different. In many texts, the author simply neglects the "alpha" factor in the eqn because it is close to unity, around 0.98 to 0.998. The eqn you quote is E-M. The coefficient of the exponential is "alpha*Ies". The omission of alpha is simply an acknowledgement that its value is close to 1. Same eqn, same coefficient. No brainer.

Sorry, the equation I wrote is, once again, a different method of calculating collector current. The coefficient is different. I spent the entire morning yesterday going back over this just to make sure of what I am saying. It's a very common method of analysis.

Different equation/different coefficient. yeah, no brainer.

EDIT: I"m glad we've had this discussion. It forced me to go back over stuff I haven't looked at in decades.

 

[BrownOut]

I was looking to hear some other examples besides the voice and speaker case.
It makes it much more interesting to learn about all these types of amplifiers when you treat the load not just as a resistor nor the input as a sine wave.

Sources: Photo detector, antenna, thermocouple, etc.

Loads: motor windings, indicator light (ie - led), another amplifier stage, etc.

 

[alphacat]

Thank you very much

 

[BrownOut]

If you're not confused enough yet, look at a hartley or colpitts oscillator. The feedback network is both the source and the load!

 

[MrAl]

Who is WE? I explained clearly that a bjt amplifies BOTH current & voltage, and that the gains are given by the 3 eqns I posted. I've acknowledged all 3 eqns w/ inputs of Ib, Vbe, & Ie. I stated emphatically that all 3 variables are important to bjt operation.

The conflict was over my eqn 2) which had an alpha factor, vs. another poster's eqn similar to my no. 2), but w/o the alpha. Ic is always alpha*Ie. Many texts approximate alpha as unity, which is pretty close at 0.99, and omit alpha in the eqn. 2). I simply stated that alpha is in the eqn but close to unity. To say that Ic = Ies*exp( ) is very close to correct, the error being 1 or 2%. That is my point.

No conflict at all. Neglecting alpha will produce an error of less than 2% typical. As long as the bjt has existed, that has been the case. I've read Drs. Ebers & Molls original 1954 IEEE paper several times, and little has changed except that Drs. Gummel & Poon refined the E-M model.

I just meant that this started to turn into a discussion of whether or
not Ies=Ics or not, when in some forums they are equal and in others
they are not.
I didnt meant that statement as an end all of all end's all either, so
if you wish to keep 'discussing' this thing that actually has two answers,
then please go ahead. I think it's somewhat interesting anyway.

I have to respond to another post too but i need to reply again.

 

[MrAl]

I was looking to hear some other examples besides the voice and speaker case.
It makes it much more interesting to learn about all these types of amplifiers when you treat the load not just as a resistor nor the input as a sine wave.

Hi again,


I think you are getting something from this thread and that's good.
I just want to offer another point of view...

Think about this:
You can make an amplifier from either a voltage dependent current
source (VDCS) *OR* a current dependent current source (CDCS), both of
which will work. To amplify a voltage with the VDCS you only need apply
the input voltage, then convert the output to a voltage with a resistor or
in the load itself. To amplify a voltage with a CDCS, you first need to
convert the input voltage to a current (with say a resistor) and then
convert the output to a voltage also with either a resistor or in the
load itself.

When the transistor is looked at as an VDCS we think about it in
those terms and design accordingly. When it is looked at as a CDCS we
think about it differently and use different equations.

For example, it's much faster to use Ic=B*Ib when we want to drive
a relay that requires 100ma and we know the min gain of the transistor
is 100...we know we can get by with only 1 or 2ma base current.
In this case, it would have been much more abstract to think about it
in terms of what the base voltage is doing while the collector current
changes...almost a waste of time. We'd have to calculate the base
current anyway.

 

[BrownOut]

Transistors can be used to provide voltage gain, current gain or both ( or neither ) It can be used as an "impeadance transformer" to accept a signal with high source impeadance and deliver into a low impeadance load. It can be used as a switch. It can be used to transform a signal to one of a different shape. It can be used as a muliplier or as a voltage reference. Sometimes, the "input" might not even be electrical, as in the case of thermal input.

The transistor is one of the most amazingly versital devices ever invented!

 

[Claude Abraham]

I just meant that this started to turn into a discussion of whether or
not Ies=Ics or not, when in some forums they are equal and in others
they are not.
I didnt meant that statement as an end all of all end's all either, so
if you wish to keep 'discussing' this thing that actually has two answers,
then please go ahead. I think it's somewhat interesting anyway.

I have to respond to another post too but i need to reply again.

Ok, 1 point at a time is fair. Regarding Ics & Ies, they are NOT equal in general. The collector region is doped lighter than the emitter. So the b-c jcn has a different "Is" reverse sat current than the b-e jcn. In Shockley's diode eqn, Id = Is*exp( ), the "Is" is that scaling current, the reverse sat value. In a diode there is but 1 junction, so "Is" describes the scaling current for that 1 junction.

In a bjt, "Is" differs because there are 2 junctions not just 1 like the diode case. Since the doping densities differ, we must clearly differentiate betwen Ics & Ies, the reverse sat currents, or scaling currents if you prefer, for the 2 junctions, c-b, & b-e.

Regarding alpha, every text which *derives* the E-M relation has no alpha in the Ie eqn, and includes alpha in the Ic eqn. So, Ies*exp( ) is the emitter current. The collector current is alpha*Ies*exp( ).

Any text which presents Ies*exp( ) as collector current is simply rounding alpha to 1, a good approximation. If "Is" is used instead of "Ies", then it is *understood* that "Is" is referenced to the *b-e* junction, not the b-c junction. Since bjt's are usually operated in the active region, this is understood, since the b-c junction is reverse biased. When used as a saturated switch, however, both Ics & Ies come into play since BOTH the b-c & b-e junctions are forward biased. You cannot just use "Is" in that case. You must acknowledge the scale currents for both junctions, seldom equal due to differing doping levels. When you say "in some forums they're equal & in some they aren't", well, whatever.

Happy 4th of July to all.

 

[alphacat]

Hi again,


I think you are getting something from this thread and that's good.
I just want to offer another point of view...

Think about this:
You can make an amplifier from either a voltage dependent current
source (VDCS) *OR* a current dependent current source (CDCS), both of
which will work. To amplify a voltage with the VDCS you only need apply
the input voltage, then convert the output to a voltage with a resistor or
in the load itself. To amplify a voltage with a CDCS, you first need to
convert the input voltage to a current (with say a resistor) and then
convert the output to a voltage also with either a resistor or in the
load itself.

When the transistor is looked at as an VDCS we think about it in
those terms and design accordingly. When it is looked at as a CDCS we
think about it differently and use different equations.

For example, it's much faster to use Ic=B*Ib when we want to drive
a relay that requires 100ma and we know the min gain of the transistor
is 100...we know we can get by with only 1 or 2ma base current.
In this case, it would have been much more abstract to think about it
in terms of what the base voltage is doing while the collector current
changes...almost a waste of time. We'd have to calculate the base
current anyway.

In the case of the relay, wouldn't it be more wise to have the transistor to act as a switch rather than as an amplifier?
For example, we used to work with a 5V @ 80mA relay, and since we wanted the relay driver to have low power dissipation, we used a 5V power source (VCC) and a BJT switch, that way the BJT only dissipated 80mA*VCE_SAT.
If we would have used it as an amplifier, we would have needed to forward bias it, first need to use around a 7.5V power source, and the BJT's power dissipation would be around 2.5V*80mA (2.5V is the VCE when the BJT is forward biased, and is found as closer as possible to the center of the high gain region).