Be very careful not to get confused by thinking that the OP's equations for collector current is the same as equation 2 stated without alpha. His equation calcualtes collector current directly as an approximation of base diffusion current. The constant in his equation is abbreviated as "Is":

IC = Is*e^vbe/vt
IE = IC/α

That isn't the same equation as (2) given above and ignoring alpha.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Is, or Ies, same thing. The "Is" notation is used when describing diodes. There is only 1 junction in a diode, 2 terminals. A bjt has 2 junctions, b-c, & b-e. Since the emitter & collector regions have different doping levels, they have differing scaling currents. The Ebers-Moll equations specify "Ics" for the c-b scale current, & "Ies" for that of the b-e junction. I am not the least bit "confused". The alpha factor cannot be omitted. The collector current depends on alpha all the time for any bjt.

If the base region was so wide that no electrons (npn polarity) make it through the base to the collector, the equation Ic = alpha*Ies*exp((Vbe/Vt)-1), still holds. The result is zero Ic, since alpha would be zero. Without the alpha factor, the Ic value predicted by the equation is wrong. You can never omit alpha. The exponential without alpha is the emitter current.

Collector current is simply the emitter current times alpha. That is the most basic relation. Of course, Ie cannot be established w/o Ib & Vbe. Ib & Vbe are indispensable since Ie couldn't exist w/o them. Hence eqns 1) & 3) account for the functional relations involving Ib & Vbe.

All 3 eqns are important. All 3 variables, Ib, Vbe, & Ie, are indispensable and all necessary to produce Ic. The OP question was if the bjt is amplifying current or voltage. I answered that it amplifies BOTH. No other answer remotely makes any sense at all. Eqn 1) describes the bjt current gain, & eqn 2) its voltage gain. Eqn 3) describes the physics of bjt action at the individual charge carrier level. This is well known for over half a century. No one has ever successfully challenged this well established bjt model. BR.

__________________
Claude

 

First of all, Is is not the same as Ics. Neither is it the same as the diode coefficient.

Secondly, the equation that calculates collector directly does not come from the ebbers-moll model. Instead, it uses the difussion profile in the base as an approximation to collector current. It's a very common way to calcualte collector current as a function of base-emitter voltage. But it's not ebbers-moll

Thirdly, I'm not concerned about your confusion. I'm concerned about the student that's learning transistors, and hoping that he doesn't get confused into thinking his equation is the same as eqn2 and ignoring alpha.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

As far as "Is & Ics & Ies" go, I don't follow. In a diode, Id = Is*exp((Vd/Vt)-1). "Is" is simply the scaling current of the p-n junction, which depends on temperature strongly, dimensions of the junction, doping profile, etc.

In a bjt, "Is" is usually understood to be "Ies". In the forward active region of operation, the b-c junction is reverse biased so that "Ics" is small compared to Ic. Hence Ic = alpha*Ies*exp((Vbe/Vt)-1). Often, rather than write "Ies", some just use "Is". It is the scale current of the b-e junction. Eqn 2) is indeed the Ebers-Moll eqn. Of course, eqn 1), Ic = beta*Ib, is NOT Ebers-Moll, nor is eqn 3).

"Is" & my "Ies" are simply the scaling current, or "reverse saturation current" of the b-e junction. They are the same thing. Look it up.

Also, you defined Ic as Is*exp( ), then Ie as Ic/a. That is not correct. Ie = Is*exp( ), & Ic = alpha*Ie. First the b-e junction is forward biased. Ib, Vbe, & Ie are established. The holes from the base move towards the emitter, the electrons move from emitter to base, & Vbe is present. No Ic exists at the start. Then a moment later, electrons emitted from the emitter transit through the base & continue into the collector. But a few holes from the base enter the emitter, & a few electrons in the base recombine & never reach the collector. The holes frpm base to emitter & the electrons from the emitter recombined in the base do not contribute to Ic. Thus alpha is defined as Ic/Ie. Ie comes first, then a moment later Ic is established. Ic is almost equal to Ie for a good device, not quite equal.

This is well established. Ie produces Ic, not vice-versa. Ie comes first along with Ib & Vbe. Then Ic follows. Ic is defined by alpha*Ie at the device level.

__________________
Claude

 

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

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?

__________________
Claude

 

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

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.

__________________
Claude

 

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.

__________________
Claude

 

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.

 

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

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.

 

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.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.