# Common Collector Amplifier - Power Supply confusion

Discussion in 'General Electronics Chat' started by timkoupe, May 3, 2013.

1. ### timkoupe Thread Starter New Member

May 2, 2013
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I may be getting into the author's material not quite fleshed out. I've been all through this awesome online text book, DC and then AC, and now going through semi-conductors and I've thoroughly enjoyed how well the author seems to empathize and predict the details and questions that pop in my head as I'm reading. But this amplifier chapter is leaving some holes in the theory, or else I'm just really dense when it comes to transistors.

On the common collector amplifier the author explains that Vout will nearly be equal to Vin, minus the .7 volt drop. Ok, that makes sense to me *only* if you pretend like that power supply off to the right doesn't exist. Why doesn't that voltage add to the voltage being dropped from Vin across R-load?

In fact, I don't see any reference to the supply voltage on this chapter at all. It's as if he forgot about it, and by extension forgot to explain what happens to it, and why I don't see it across R-load.

Is there some other online resource I should be using?

2. ### Ron H AAC Fanatic!

Apr 14, 2005
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Vout will indeed equal to Vin - ≈0.7V. The remainder of the supply voltage (call it Vcc), will be across the transistor (Vce). Be aware that Vcc must be greater than or equal to Vin for this to happen.
You are correct in saying that Vout will be about the same with or without Vcc, but there will be no current gain in this case.

3. ### #12 Expert

Nov 30, 2010
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I'm with Ron, but I felt I had to say, "good call". The "output" of that transistor will be about Vin - .7V whether the supply on the right exists or not. The voltage from the right does not add to the emitter voltage because the transistor is stopping it from doing that. If the transistor allowed too much current to flow into Rload from the collector supply, the voltage developed across Rload would approach the voltage of Vin and shut the transistor off. Thus, it becomes a self regulating transistor circuit, which is the whole point of why this circuit was invented.

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4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I think that in CC amplifier we don't care about Vcc much, because BJT base-emitter junction behaves just like a diode.
And collector-emitter behaves like a constant current source.
And as we all know voltage drop across BE junction ( a diode) is equal 0.6V....0.7V.
And this is why emitter voltage is equal to
Ve = Vin - Vbe = Vin - 0.6V

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5. ### Ron H AAC Fanatic!

Apr 14, 2005
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To get current gain, Vcc needs to be >= Vin. (Strictly speaking, you can still get current gain with Vce<Vbe, but that is a fine point).

6. ### Ramussons Active Member

May 3, 2013
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True as far as DC input is concerned.

What happens if the input is a Sine Wave of amplitude A?
Will the output amplitude be (A - 0.7)?

As far as I know, the Common Collector, also known as a Emitter Follower, only does a DC offset of 0.7 V. The Amplitude will still be A.

Ramesh

Last edited: May 6, 2013
7. ### Ron H AAC Fanatic!

Apr 14, 2005
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That's correct. The DC offset is ≈0.7V (negative for NPNs, positive for PNPs). The AC gain is almost unity, because of the fact that the offset is nearly constant, independent of DC level, over a limited range.

8. ### timkoupe Thread Starter New Member

May 2, 2013
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Ok, it almost sounds like we are violating ohm's law. Surely not.

If Vin = 1.5 volts, then Vout = 1.5V - .7 = .8 volts.

So we are saying that we are dropping .8 volts across R-load? And if so, then how do we get current gain of more than 160 uA through a fixed value 5K resistor? In other words...how do you increase current through a 5K resistor without increasing voltage?

With an increase in current, we must get an increase in voltage or Ohm's law gets reduced to a suggestion, right? (Just kidding of course...it's been a long weekend.)

E/I = R. I am not seeing it...

Thanks for your help by the way...

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But who said that we increase current through a 5K resistor?
The situation looks like this

And as you can see, emitter current is equal to Ie = 160μA
But the base current Ib is (β + 1 = 101) times smaller then the emitter current.
And our 1.5V input voltage source "see" our load resistance (Re resistor) not just as 5kΩ resistor. But Vin see much higher resistance.
So as you can see in CC amplifier the transistor current "gain is use" to increase resistance seen by input voltage source. Vin is not loaded by 160μA. Vin load current is (β + 1) times smaller than the load current.

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10. ### Ramussons Active Member

May 3, 2013
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The addon to this is Ie, the Emitter Current Ie is independent of B2, implying that Ic, the Collector current too is independent of B2. This further can be enhanced to what we call a "(Constant) Current Sink" or a (Constant) Current Source for a PNP Transistor.

Ramesh

11. ### timkoupe Thread Starter New Member

May 2, 2013
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You've said something here, bolded (emphasis mine), that makes me think it's at the crux of my problem understanding this circuit.

Both of the batteries (B1 and B2) seem to add to the voltage across the load resistor. That's what makes this so hard for me to get....at saturation, the emitter to collector is like a short, right? ( a current controlled switch) And the load resistor is connected between the collector lead and emitter lead, so why isn't that B2 voltage being dropped across the resistor *in addition* to the B1, input voltage?

In other words you say if too much current flowed into the load resistor from the collector supply, the voltage across that load resistor would approach the voltage of Vin and shut the transister off....how is that? I think that if I *see* that conceptually, I'll get this.

At this point, thanks to everyone in this thread, I can analyze the circuit and recognize a CC amplifier, or emitter follower as it may be called. But I don't really *see* it. You know what I mean?

So yeah, Vin = 1.5 volts with .7 volts dropped across the Vbe junction, leaving .8 volts across the 5K load = 160 uA. 160uA/B+1 = 160uA/100+1 = 1.584 uA base current, and then 158.4 uA collector current with B=100. Emitter current = base and collector so 160uA.

I can do that...but I'm not "seeing" the operation.

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If we assume that we have BJT at saturation (emitter to collector is like a short).
We have a 10V at emitter (across Re resistor). And we have 1.5V at base.
So Vbe = Vb - Ve = 1.5V - 10V = -8.5V so the transistor is cut-off.
Voltage at emitter cannot be greater than the voltage at base. Because Vbe is less then 0.6V and the BJT will be in cut-off.

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13. ### Ron H AAC Fanatic!

Apr 14, 2005
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Here is a simulation. The collector voltage is fixed at 10V. The base voltage is swept from 0 to 10V.

Plotted are the base current (top), collector and emitter current (middle), and base and emitter voltages (bottom). Note that collector and emitter currents are almost equal.

Study this and see if it adds to your understanding. Come back with questions.

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14. ### timkoupe Thread Starter New Member

May 2, 2013
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Ok, ok...I think I get it...

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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We can also use a circuit theory

If we assume BJT current gain equal to β = Hfe = 99.
We can use II Kirchhoff Law and write

Vin = 0.6V + Ve

So

Ve = Vout = Vin - 0.6V As you can see just by using II Kirchhoff law we can tell that B2 voltage source don't have any influence on Ve voltage.

Ve = 1.5V - 0.6V = 0.9V.

Next we can find emitter current just by using Ohm's law

Ie = Ve/Re = 180μA.

And form the first Kirchhoff's law we have

Ie = Ib + Ic = Ib + Ib*β = (β + 1)*Ib

And from here

Ib = Ie/(β + 1) = 180μA/100 = 1.8μA

And Ic = Ib*β = 178.2μA

And again we use second Kirchhoff's law to find Vce.

Vb2 = Vce + Ve
----> Vce = Vb2 - Ve = 9.1V

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16. ### timkoupe Thread Starter New Member

May 2, 2013
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The key to my problem with this circuit was my total misapplication of input and supply voltages.

Out of frustration, I just drew this out and a tech friend of mine practically slapped me across the face.

In my mind, I was seeing those the voltage from those two batteries "adding", as if 11.5 volts was going to drop across that resistor. I don't know why I did that since those voltages are in parallel. *Currents* would add, not voltages.

That is basic DC electronics, and I just completely blew it.

This is why I kept saying that I couldn't understand why the supply voltage was not adding to the voltage of the load resistor for this common-collector, emitter follower amplifier example. I wasn't realizing that at saturation when the collector to emitter was an essential short, at best we have two batteries in parallel to the load.

Really there's more to it than that, but clearly two batteries in parallel do not add voltages, only current capacity.

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17. ### Ron H AAC Fanatic!

Apr 14, 2005
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And if they are not the same voltage, there will be large amounts of heat generated.
Parallel batteries will, by definition be at the same voltage.