# Common-collector confusion

Discussion in 'General Electronics Chat' started by halfgaar, May 15, 2011.

1. ### halfgaar Thread Starter New Member

May 15, 2011
4
0
Hi,

I'm studying the common collector configuration explained on the site. I'm confused about the current gain of the circuit.

Let's take the image from the article:

I don't understand why there is any current gain at all. I mean, when you connect a load to Vout (node 0 and 3), the Rload is effectively decreased. Doesn't this proportionally increase base current?

If Vin=5V and the new load impedance (Rload2) is 10R, base current would be 0.5A. And because the voltage over the output load, in this case Rload2, is always 4.3V, how can the current through Rload2 be anything else than 0.5A, equaling base current?

2. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
298
Most of the emitter current actually comes from the collector, and only a small fraction comes from the base.

Suppose Rload2 draws 0.5A, and we have a transistor with current gain β = 50.

Ic = β * Ib, and Ie = (β + 1) * Ib, so Ib = Ie/(β + 1). The increase in base current will be 0.5A/51, just under 10mA.

3. ### halfgaar Thread Starter New Member

May 15, 2011
4
0
I guess that's where I'm lost.

I drew in the new load I was talking about:

Without the red load, it seems to me base current would be:

$
I=\frac{V}{R} = \frac{4.7}{5000} = 0.00094A
$

And with the new load:

$
I=\frac{V}{R} = \frac{4.7}{9.98} = 0.47A
$

But if you're saying most of the emitter current comes from the collector, there must be some other factor limiting/determining the base current, aside from the emitter resistance.

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
That other factor is the beta of the transistor designated with the symbol β.

hgmjr

5. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
There is, it is beta.

Ib ≈ Ie / β

This is why the other name for this circuit is a voltage follower. All of the gain is expressed as current, not voltage.

I assume 10R = 10Ω.

6. ### halfgaar Thread Starter New Member

May 15, 2011
4
0
I think I realize my error. I thought the Hfe (β) was just a multiplication factor that determines how much collector current will flow based on the base current. But does it also work the other way around: does it also determine the amount of base current based on how much collector current flows?

7. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
298
If we know that two currents are in a certain ratio, (in this case 1 : β) then knowing either current we can always find the other.

If the circuit conditions put the transistor in the active region, IC = β * IB , and by simple algebra we get IB = IC

Edit: You cannot really determine the base current from the collector current if the transistor is saturated, a condition where a relatively large base current is supplied, so that the collector-emitter voltage is reduced to a low value.

In this case a lower value of HFE is said to apply (HFE(SAT) ) to calculate the required base current to ensure saturation, but such operation is not in the active region and the base and collector currents are not linearly related.

Last edited: May 15, 2011
8. ### Audioguru New Member

Dec 20, 2007
9,411
895
A silicon transistor has a base-emitter voltage of about 0.7V, not only 0.3V which is for an old geranium transistor.

So with +5.0V on the base then the emitter voltage is about 4.3V, not 4.7V as you showed.

EDIT: Sorry it is spring time and I have flowers on my mind. I planted some red geraniums today. The old transistor was made from germanium.

Last edited: May 15, 2011
9. ### halfgaar Thread Starter New Member

May 15, 2011
4
0
Sorry, the 4.7 was basically a typo. I knew about the 0.7V drop.

Anyway, I think I get it now. Thanks.

10. ### Rbeckett Member

Sep 3, 2010
205
32
AG even when your mind is else where your still way smarter on this stuff than me and I appreciate you ongoing efforts to help new learners like myself and all the others. I bet a geranium diode will at least smell good if you let the magic smoke out... Hee Hee, sorry could't help myself..
Bob