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?

 

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?

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.

 

Most of the emitter current actually comes from the collector, and only a small fraction comes from the base.

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.

 

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.

That other factor is the beta of the transistor designated with the symbol β.

hgmjr

 

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.

 

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?

 

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 /β

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.

 

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.

 

Sorry, the 4.7 was basically a typo. I knew about the 0.7V drop.

Anyway, I think I get it now. Thanks.

 

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.

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