# Common Emitter-Base Amplifier - Inversion confusion

Discussion in 'General Electronics Chat' started by timkoupe, May 2, 2013.

1. ### timkoupe Thread Starter New Member

May 2, 2013
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I can't seem to find an answer to this question that is burning my brain...

In the Common-Emitter BJT amplifier circuit...the collector output voltage is inverted. That makes sense to me, sort of, as the base current increases the voltage between the emitter and collector decreases. But the current from emitter to collector *also* increases, right along with the base current...so why is the collector current not out of phase with the collector voltage 180 degrees?

I would expect the current from emitter to collector and emitter to base to be in phase with each other. And the voltage from emitter to collector to be 180 degress out of phase.

In fact, I'm thinking if you drop the amplifier out of this altogether and put two resistors in series, R1 + R2 with a fixed voltage source V1, and varied the resistance of R1, oscillating back and forth, you would also notice the current out of phase with the voltage *across R1*. As R1 increases, the voltage drop increases, and the current decreases (due to total circuit current decreasing), and as R1 decreases, the voltage drop decreases, and the current increases.

Am I wrong on this? Very confused...

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2. ### nigelwright7557 Senior Member

May 10, 2008
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There is no capacitor or inductor there so the voltage and current are in phase.

3. ### MrChips Moderator

Oct 2, 2009
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... because the load at the collector is connected between the collector and the positive supply.
You need to measure the voltage across the load, not just at the collector.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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When base current increases the collector current also increases.
So voltage drop across Rc resistor must also increase (Ohm's law VRc = Ic * Rc).
So VRc increases the Vce voltage must drop. Because V_Supply = VRc + Vce.
As you can see when input voltage is increasing ( base current also) the Vce voltage decrease. So we have 180 degress out of phase.

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5. ### timkoupe Thread Starter New Member

May 2, 2013
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Ok, that's my problem...not considering the load resistor. As the voltage across collector to emitter decreases, it is increasing across the load resistor, right along with the collector current. Not sure why that wasn't obvious before...but thanks for pointing that out.

Ok...so that brings up another question. Using All About Circuits example from that chapter...the author carefully chooses B=100, 30uA current source at the base, 15V emitter-collector source and 5K load at R1 - which would give 3mA at R1 at full saturation. That matches the gain of 100 for the transistor example.

So what happens if the emitter-collector source voltage was 100V? With a 30uA current source at the base, the max current out of the collector would still be 3mA, which would still drop 15V across R1....what happens to the other 85 volts?

By the way, I really appreciate the replies, you guys are quick and helpful. Thank you!

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well we have 15V across Rc resistor so the rest of the supply voltage ( 100V - 15V = 85V ) will be present between collector and emitter (ground).

7. ### timkoupe Thread Starter New Member

May 2, 2013
16
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Ok I'm back to not seeing the inversion then.

As base current increases, collector current increases, and R-load current increases. In terms of voltage, as base current increases, emitter to collector voltage decreases but voltage across R-load increases.

I don't see any inverted output except the voltage across emitter-collector. All currents increase and decrease in unison, and voltage across R-load increases in unison with the currents.

Clearly there is some key point I am missing...

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But what about Vin vs Vce voltage ? You don't see inversion here?

9. ### timkoupe Thread Starter New Member

May 2, 2013
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Yes, I absolutely do. That's what I meant by "I don't see any inverted output except the voltage across emitter-collector".

Is that the only inversion related to this circuit? I thought the inversion was supposed to appear at R-load as well. If not, then I think I understand this better than I thought.

So if Vin vs Vce is inverted, and Iin vs Ice is not, then wouldn't that mean that Vce and Ice are 180 degress out of phase? That was my original question in the OP.

Thanks for your help and patience Jony130...

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well you are right. And normally we connect load resistance parallel to collector-emitter junction

So Vout is also 180 degress out of phase.

Yes, Vce and Ic are 180 degress out of phase. But we never look it that way.

11. ### timkoupe Thread Starter New Member

May 2, 2013
16
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Which part am I right about...that Vr-load is inverted right along with Vce, or that *only* Vce is inverted and Vr-load is not? (and is Vr-load the same as Vout?)

Because now I'm on the Common-collector amplifier and the first point he makes is that Vout is not inverted. Again, I do not see any difference with Vce here just because the load has been moved. Vce is *still* inverted, it would seem. As base current increases, Vce decreases, changing from an open circuit to a throttled path..still...

That tells me there is still something about the common-emitter Vout that I am not getting. If you increase current through Rload, and Rload is a fixed value, then voltage *is* going to increase right along with the current.

I don't see any inversion except Vce, and in that case, how is that any different with common-collector then? Keep in mind I'm still looking at the simplified circuit with a single resistor, R-load, a single battery Vin and a single battery V1. I'm not ready for all the bias resistors and caps in the pic you posted above.

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If our output voltage is taken from collector the Vce is inverted right along with Vr-load. If we connect load between the collector and ground.
In Common-collector amplifier the output is taken from emitter. So output voltage is present between emitter and ground.
Read this ( Ve = output voltage)

And in common-emitter amplifier the output voltage is taken from collector and ground

Last edited: May 3, 2013
13. ### timkoupe Thread Starter New Member

May 2, 2013
16
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Ok...here's my circuit...

Now...as I1 increases, Ic increases, which means Ir-load increases, which means Vr-load increases. It has to, because the R value is fixed.

If I1 = 0, then Ic = 0, Ir-load = 0 and Vr-load = 0. Vce = the full 15V since nothing is being dropped across the load resistor. So far Vce is inverted, but Vr-load is right in phase with the I1, base current.

When I1 = 30uA, then Ic = 3mA, Ir-load = 3mA, Vr-load = 15V. Vce = 0V since we have saturation. All of this is in phase with I1. Everything is increasing right along with I1 except Vce.

How is that wrong? How could Vr-load possibly be inverted?

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Vr-load is in phase with Vin. But Vout is inverted (voltage between collector and the ground).

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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And normally this 5K resistor is not treated as a load resistor. Because our Vout is Vce. So we connect RL between collector and gnd

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16. ### timkoupe Thread Starter New Member

May 2, 2013
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Ok, that makes sense then.

Although I'm not sure what the purpose of the observation is since there is no load or components between collector and ground, it is the same as Vce.

Likewise, Vce is just as inverted for the common-collector circuit.

Now, if we're looking at Vout from collector to ground, it is not inverted since our load resistor is in there. But Vce is *still* inverted.

I guess I'm not sure about the purpose of the distinction of invertion since it has nothing to do with the load or any other components, and the same condition exists in both amplifier arrangements.

Put another way...the moving goal posts don't make any sense. Why are we looking at voltage from collector to ground in common-emitter? Of course it's inverted, but there is nothing there. The load resistor is where the energy is going and it's right in phase.

And when we are looking at common-collector, we arbitrarily move the goal posts to looking at voltage from emitter to ground, only this time we are looking at the load resistor and observing it's in phase. Well hell, the load resistor is in phase with the input in both designs. Why is this a point of distinction?

Maybe the distinction has relevance after further study?

17. ### timkoupe Thread Starter New Member

May 2, 2013
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Ah, that helps explain my "distinction" issue, ha ha. Thanks!

18. ### Jony130 AAC Fanatic!

Feb 17, 2009
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There is a huge difference between common-emitter (CE) and common-collector (CC) amplifier.
In CE amplifier Vout is larger the the Vin. So we have a voltage gain amplifier.
And we have this amplification thank to this 5K resistor connect between collector and Vcc. Simply this 5K resistor converts Ic current to a voltage.
The large Rc resistance the large the voltage gain. Because we need a relatively small changes in the base current. To change the collector voltage "quicker" from Vcc to saturation.

And in CC amplifier we don't have any voltage gain and we don't care about Vce voltage.

19. ### BillB3857 Senior Member

Feb 28, 2009
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If you look at the collector/emitter connection of the transistor as a variable resistor that is in series with the collector load resistor, I think it may clear things up. As the transistor turns on, the effective resistance decreases. Overall current goes up, but the point equivalent to the collector will see a decrease in voltage. Simple voltage divider theory.

20. ### timkoupe Thread Starter New Member

May 2, 2013
16
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Exactly, that's exactly what I typed out in the OP:

The point that comes to mind, of course, is the point equivalent to the collector seeing a decrease in voltage seems pointless, because there is no load or any other components there. It is, essentially, Vce. And Vce is *always* inverted with respect to the input, even in the common-collector.

I just figured there will be a good reason for that later (like the one Jony provided). Thus far, the observation that collector to ground is inverted is equal to pointing out that Vce is inverted.

I'm one of those people that really gets stuck on details and find it powerless to move forward when that happens.