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Common collector configuraton

So if beta = 300, thrice the current will exist (Emitter current), but how?

No, it won't. Last time I gave an example where the current in the emitter circuit was 10 ma. That current was "locked" into its value because of the 5 volts on the base terminal, the Vbe, the current limiting resistor, the the voltage across the LED. If you used a transistor with a 50,100,300,400,or 500 beta, that emitter current (10 ma) would stay the same, because beta does not affect the emitter current in this circuit. The only difference would be is that the waste current in the base circuit would be less at the higher betas. But who cares what the waste current in the base is? It does not control the emitter current.

I only knew few equations that's why I went through all this steps to connect beta and Ie. but from your reply "You'r lost at sea" I guess I'm wrong.

That is correct. You still are not getting the "big picture". You don't have to worry about beta for the reason I explained above.

So please clarify how it gets changed depending on gain?? Is there any equation??

How does what get changed? Why are you looking for an equation? I already showed you how to calculate the current limiting resistor needed for the LED.

One more Is there any current flowing through collector ( current taken from supply connected to Collector??

Of course there is. That is the main current source for the LED.

Or is it only flowing through base - emitter circuit?

The current present in the base is the unavoidable waste current that ensues whenever a control voltage (5 volts in this case) is applied to the base terminal. It is usually very small compared to the emitter/collector current.

Ratch
 
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May I point to the fact, that Vbe does not "drift" with rising temperatures? Instead, it is the collector current Ic that increases with the temperature - and we have to reduce Vbe externally by app. 2mV/K to bring Ic back to its former value.
So - where is the advantage to view the transistor as a current amplifier?

Remember: The tempco is -2mV/K for Ic=constant.

I meant to say

if temperature drifts higher using a true voltage source, you must allow for the Shockley (NTC) effect in the transconductance model , when it is far simpler to use the Norton equivalent circuit with an external base R and assume a range of current gains provided in the datasheet for both linear Vce and Vce(sat).

But I was trying to simplify my opinion, which has been successful in my design career since 1975.

I have never used the transconductance model in a formula in hundreds of designs.

But I have used thermal compensation between crossover bias and output drivers, for example using thermally connected parts.

But you can use the transconductance formula if you choose with an ideal voltage source.
 
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Harald Kapp

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'hfe' is just one parameter out of four of the hybrid or h-parameter bjt model. To fully undestand the importance and the working of hfe you neeed to understand the full model, not only a single parameter.

The full model (see link for image) specifies:
I2 = h21*I1 + h22*V2
with
h21 = hfe
I1 = Ib
I2 = Ic
as a special case (details again see link).

Using Ic = hfe*Ib is a simplification of the above term neglecting the influence of h22. Plus there is the term
V1 = h11I1 +h12V2
with
V1 = Vbe
etc. which is completely neglected.

So if beta = 300, thrice the current will exist (emitter current), but how?
beta=300 is 300 times, not 3 times.
This current will not necessarily flow. The equation gives you the max. current that can flow provided the external circuit (power supply, resistors etc.) provides enough power.
I like the analogy to a water pipe: When you open the faucet, you allow a certain amount of water to flow. The wider you open the faucet (read: more base current to the transistor), the more water may flow out of the faucet (read: collector current through the transistor). If, however, the flow of water is limited by e.g. a very thin pipe (read: high resistance), the amount of water flowing out of the faucet is limited by the amount of water that can pass through the thin pipe, regardless of your opening the faucet very wide.

Note that the h-parameter model is valid for small signals only. This means that you have to set the DC operating point of your circuit by suitably selecting and dimensioning a bias circuit.
Look at a typical transistor datasheet, e.g. for a 2N2222.
In the section 'SMALL−SIGNAL CHARACTERISTICS' you'll find 'Small−Signal Current Gain' hfe varying with collector current.
In the section 'ON CHARACTERISTICS' you'll find 'DC Current Gain' HFE (note the fine difference in the subscript which is capitalized!), again varying with collector current.
You'll also find that hfe and HFE per se have a very wide range. This is why any reliable circuit needs to incorporate feedback and possibly trimming to compensate for this variety. Both min and max values have to be taken into consideration.
 
@Athul
When using emitter followers ( preferred name over CC )

we think of them as Voltage Followers with a diode drop, as long as there is sufficient current in the base to "saturate" the base emitter using the minimum current gain.at specified Vce.

This results in an output impedance at the emitter which is Rb/hFE (neglecting resistance in the Base-emitter diode. This makes it more of a voltage source thus a current amplifier.

so current must be limited by an external resistance in the emitter to ground is needed plus LED voltage drop.
If we don't limit the emitter current, it will be amplified by hFE
Ok?

In retrospect, the hybrid model is excellent for detailed math analysis, but often is not needed to solved simple common designs and as such these parameters are not shown in datasheets as "h21" , rather it is shown as hFE or beta.
 
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I have never used the transconductance model in a formula in hundreds of designs.

But you can use the transconductance formula if you choose with an ideal voltage source.

Perhaps I couldn`t express myself clear enough - I didn`t speak about any "model". I was referring to the actual behaviour of the transistor when it is biased - as it is common practice in most amplifier applications - with a constant voltage at the base (voltage divider). In this case, we are using an emitter resistor Re (emitter degeneration) for stabilizing the qiescent collector current Ic against temperature deviations and uncertainties (tolerances) of the Ic/Ib ratio. As you know, this works because an unwanted increase in Ic results in a corresponding Vbe reduction.

I think, in this case, we do nothing else than to exploit the transconductance properties of the BJT (possibly without realizing this fact), don`t we?
 
when driving linear loads , agreed.
but not when driving diodes (LED)
Then I consider base current must over driven with suitable gain reductions for Beta to ensure saturation and should use current limiting R. in either case CE or CC mode.
But inverted CE is preferred with potentially lower drop than Vbe.
 
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