CC amplifier configuration.

Hey,
I ran into this configuratuin of CC amplifier (output is taken from Emitter) and wanted to know what is the advantage of using such configuration?

The input resistance of this amplifier is approximately RE||RL, which in normal CC configuration, its β times larger: β*(RE||RL).

Since CC amplifier's gain should be as close to unity as possible, then i dont understand why would one want to use such configuration.

Thanks.

 

Common Collector amplifiers are useful in buffering the signal from a previous amplifier stage (Typical one with fairly high output impedance) so that the load resistance does not appreciably disturb/load the output of the previous amplifier stage.

hgmjr

 

Hello,
Thank you for your answer.

In your example, lets call Rout1 the output resistance of the first stage, and Rin2 the input resistance of the second stage - the CC amplifier.

As you said, Rout1 is usually too large, and we want to decrease it, therefore we use the CC amplifier, which has a pretty low Rout.

But if Rin2 is low, and Rout1 is large, then the output voltage of the first stage will be degraded pretty badly - the attenuation factor will be Rin2 / (Rin2 + Rout1) .
So wouldnt you rather Rin2 to be as large as it can be?

(As i said, this CC configuration causes Rin2 to be β times lower than its value in the natural CC configuration, thats why i was asking what is so special about this configuration that would make us wanna use it).

(When saying natural CC configuration, i meant this

 

Sometimes you have a signal that has plenty of voltage already, but is at too high of an impedance to use directly.

An example might be pulling a radio signal from a tuned loop antenna - this signal can easily be several volts with no load on the loop, but at a very high impedance.

If you connect a relatively low-impedance input amplifier to the loop, the load will detune the loop and the signal voltage will drop (plus you'll change the frequency response of the loop), but a high-impedance input amplifier can be used.

You still might need to amplify the signal past this buffer stage, but the input impedance requirements for the second stage of amplification are much less restrictive.

 

Hey Dave,

So do you agree that the #2 CC circuit which i've uploaded - which has Rin ≈ β * (R_EMITTER || R_LOAD) - is better than the #1 CC circuit - which has Rin ≈ (R_EMITTER || R_LOAD) ?

 

Sorry, I have a general knowledge of transistor design, but I don't know enough of the engineering details to help with that question. Maybe others can help with this.

 

If you want help in understanding this, you need to be consistent in the names of your resistors. The names in your schematics do not match those in your equations.

 

The CC amplifier (emitter follower) is a current amplifier not a voltage one.
If we assume we have the input signal that has 1V and 1K internal resistance. And our load will have RL=600Ω.
So if we connect the input signal direct to the load, the voltage on the RL will be equal
UL=1V*(600Ω)/(1kΩ+600Ω)=0.375V
If we add emitter follower between the input source an the load
Then resistance that is seen by input source will by (β+1) larger then RL.
If β=99 then
RL seen be a input source:
Rin=100*600Ω=60KΩ so the voltage on a RL will by
0.983V


In a real circuit

Rin=Rb || [ (β+1)*( (Re+re) || RL ) ]


http://forum.allaboutcircuits.com/showpost.php?p=149702&postcount=13
http://forum.allaboutcircuits.com/showpost.php?p=153702&postcount=8

 

Hey Jony,

The CC amplifier has a large Rin and low Rout, therefore it would make a very bad current amplifier, but excellent voltage buffer.

 

Well maybe I should say "current booster" or " impedance transformer".
But it's true, an emitter follower has a current gain.

 

A minor correction:

Rin=Rb || [ (β+1)*(re + (Re || RL )) ]

 

Iload>>Isource. That's a current amplifier.

 

In this circuit


Ig=181uA and IL=3mA
Ki=16[A/A]
So we have the current gain, So this isn't a current amplifier?

 

It is. You just offered a concrete example of what I said.

 

Yes you're right, don't understand what you mean in your answer.

 

I said:

The symbol ">>" means "much greater than". So, Iload is much greater than Isource.
You said:

In other words, IL>>Ig. Which is equivalent to what I said.