# Volume III - Page 203

Discussion in 'General Electronics Chat' started by chal-lo, Oct 16, 2012.

1. ### chal-lo Thread Starter New Member

Sep 10, 2012
6
0
I am at the point 4.6 "The common-collector amplifier"

After figure 4.40 it says:

"...it would be reasonable to presume that this amplifier will have a very large current gain."

I don't see why an CE vs a CC amplifier will have less current gain. I have not tested it with SPICE, but with iCircuit instead, and find not significant differences.

But above all I cannot understand why does it say that. Current gain should be the same and given a certain Vbe (aprox. above 0.7V) it should reach the Beta ratio (not more under regular humidity and temperature conditions).

On the other hand, in page 205 after figure 4.43 it says:

"Given the voltage polarities across the base-emitter PN junction and the load resistor, we see that these must add together to equal the input voltage, in accordance with Kirchhoff's Voltage law."

I always analyze circuits according to the electron flow, not the conventional.
So I presume that electrons will firstly traversse the load, and then the PN junction base-emitter. If that's correct, I cannot see how possible it is that the load has a voltage 0.7V less than the input voltage. Which input voltage does it mean, the signal? (Vin) or the other DC source?, the main one.

As far as I understand, Voltage drop means the amount of voltage (energy to move electrons) that gets lost when current traverses a passive component. Which means that the voltage (according to electron flow) can't be dropped by 0.7V until it has traversed the emitter-base junction. But the load lies before the base-emitter junction.

So for this case I am completely lost.

Any guidance to clarify these concepts is greatly welcome.

Many thanks.

Chal-lo.

2. ### panic mode Senior Member

Oct 10, 2011
1,645
459
it is hard to answer your questions without pictures or original book.

In CE and CC BJT have high current gain because input current is Ib and output current is Ic or Ie which are much larger than Ib.

your "page 205" question seem to deal with CC configuration. input voltage is Vb. output voltage is Ve. there is about 0.7V difference.

there is a reason people use conventional current direction and models to explain circuits. if you are stuck on analysing every phenomena through electron flow, you will need very looong time to move forward.

3. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
I don't agree on that, on these kind of analysis the direction of current is not that important, just choose any one direction and follow it till the end of your analysis
Here we are mostly dealing with scalar quantities not vector quantities.

Remember here we are dealing with DC circuit so all the gains are also for DC
In CE configuration the DC current gain is $\frac{I_C}{I_B}$ i.e.. $h_{FE}$.In this configuration the load is connected to the collector, so the current through the load will be the current of collector i.e.. Ic.

Now in case of CC configuration the DC gain of the BJT still holds true i.e.. the $h_{FE}$ but in CC configuration the load is connected to the emitter, so the current through the emitter is the current through the load. We know that a large amount of electrons enters through emitter and move towards base and collector, a small amount (5%) of electrons go through the base and a large amount (95%) go through the collector.

In case of CE configuration the load just get the collector current Ic
But in case of CC the load gets emitter current which is $I_E = I_C + I_B$
and we know emitter current is more than collector current, hence the DC current gain of CC is more than CE.

Yes the input voltage i.e.. Vin.

It doesnt matters if the load lies before , the load is in series with the base emitter junction ,which will drop approx 0.7V and rest of the Vin will be dropped by the load as per Kirchhoff's Voltage law.

Make the circuit a bit simpler, imagine the portions Vin as voltage source, base-emitter junction as a diode and load as a resistor. Analysis the series circuit and you will understand

Good Luck

4. ### Wendy Moderator

Mar 24, 2008
21,428
2,952
I tend to agree with debjit625 on this one. Electron flow is the correct model, but it doesn't matter unless you want to truly understand the whys of things like vacuum tubes. Electrons also carry ions of material with them, lots of machines use this to coat other materials, such as sputtering machines.

There is a reason why this is the policy of AAC.

Conventional flow vs Electron Flow

A common collector amplifier, such as the one show below, has no voltage gain, but lots of current amplification. The total wattage is greater, just like with a common emitter, but the mechanism is different.

Other configurations have a little bit of both current and voltage gain, you pick which ever works best for the purpose.