Understanding Zener Diode Operation

[Femme_physics]

General basic electronics questions

I have 3 questions

QUESTION 1
--------------

http://img35.imageshack.us/img35/4011/zener.jpg

No matter what value of R I have, as long as the source voltage (16V in our case) bigger than the zener's diode voltage specification (6V in our case), the zener diode will conduct in this case (despite being backwards biased). Is that a correct assumption to make?



QUESTION 2
--------------

I'm asked if this connection is CE (Common Emitter)

http://img543.imageshack.us/img543/4290/50887855.jpg

I would say of course not, because here the arrow, which represents a diode, is pointing from the collector to the base. In common emitter, the emitter is common to both the base and the collector.

Such as:

http://img23.imageshack.us/img23/816/ce2f.jpg


QUESTION 3
--------------

Given a sinosoidal signal V(t) = 2.2 sin 314t. This signal rides on a direct current whose value is 12V. You may say that the average voltage of the signal is 12√2 V

ANSWER
----------

This one is a bit confusing to me. I want to say "NO, for a full rectified wave, the average value would be 2U(max)/pi"

 

[NascentOxygen]

Q1. You are correct.

Q2. Your justification seems confused. The emitter is a fixed terminal of the transistor device itself. The location where a circuit requires that terminal of the transistor is shown on schematics as the terminal sporting the arrow in the symbol. The emitter of a transistor is always its emitter, even when you have it wrongly soldered into a circuit you are constructing.

Q3). Try sketching the composite waveform, and graphically demonstrate its average. (Rectified? You didn't say that the question uses the word "rectified". Better be clear about that minor detail. )

 

[Femme_physics]

Q1. You are correct.

:)

Q2. Your justification seems confused. The emitter is a fixed terminal of the transistor device itself. The location where a circuit requires that terminal of the transistor is shown on schematics as the terminal sporting the arrow in the symbol. The emitter of a transistor is always its emitter, even when you have it wrongly soldered into a circuit you are constructing.

Yes, that's what I thought. :) Hmm, perhaps I didn't explain myself right. Basically, in terms of schematics, Common Emitter depends on the arrow mark on the drawing. If the arrow is pointing outside from the emitter, it's CE, any different location of the arrow would mean it's not CE. By the way-- does the arrow signify a diode? I remember my teacher saying that, although lately I lost some of the confidence in him.

Q3). Try sketching the composite waveform, and graphically demonstrate its average.

To be honest this question is a bit confusing. It's supposedly a "sinosoidal wave", but it "rides" on a direct current!? What does that mean? Which signal am I supposed to draw? My best guess is that they mean a half-rectified waveform

http://img23.imageshack.us/img23/6868/waveforma.jpg

 

[lazyaditya]

:)


Yes, that's what I thought. :) Hmm, perhaps I didn't explain myself right. Basically, in terms of schematics, Common Emitter depends on the arrow mark on the drawing. If the arrow is pointing outside from the emitter, it's CE, any different location of the arrow would mean it's not CE.


By the way-- does the arrow signify a diode? I remember my teacher saying that, although lately I lost some of the confidence in him.

Is common emitter configuration dependent on the direction of the arrow or is it common emitter only when emitter junction is common to both the input and output terminals ? can't the direction of arrow in emitter terminal depend on p-n-p or n-p-n type ?

 

[lazyaditya]

:)





To be honest this question is a bit confusing. It's supposedly a "sinosoidal wave", but it "rides" on a direct current!? What does that mean? Which signal am I supposed to draw? My best guess is that they mean a half-rectified waveform

http://img23.imageshack.us/img23/6868/waveforma.jpg

By 3 rd question diagram do you mean the waveform shown in the attachment below ?

 

[NascentOxygen]

Basically, in terms of schematics, Common Emitter depends on the arrow mark on the drawing. If the arrow is pointing outside from the emitter, it's CE, any different location of the arrow would mean it's not CE.

No, nothing of the sort. You had better review your notes on common emitter configuration.

If the arrow points in towards the base, that means its the emitter of a PNP transistor. If the emitter arrow points out, it's the emitter of an NPN transistor. The arrow always denotes the emitter.

By the way-- does the arrow signify a diode?

Were you to use just 2 of the transistor's 3 terminals, then, yes, you will find the behaviour between the base and emitter terminals to be as a diode. (So it should be, since it's a PN junction.)

I remember my teacher saying that, although lately I lost some of the confidence in him.

If he was reading this, he might reassess that 100% mark you recently scored. :uhh:

To be honest this question is a bit confusing. It's supposedly a "sinosoidal wave", but it "rides" on a direct current!? What does that mean?

You draw a sinewave symmetrical about an axis, as usual. But instead of that axis being 0 volts, you label it 12 volts and redraw your 0 volt line further down the page.

 

[Femme_physics]

If the arrow points in towards the base, that means its the emitter of a PNP transistor. If the emitter arrow points out, it's the emitter of an NPN transistor. The arrow always denotes the emitter.

I am aware of it, but we're not talking about PNP or NPN, we're talking about what is a common emitter, and in this case the emitter must point outside. So common emitter only transpires in NPN...no?

Were you to use just 2 of the transistor's 3 terminals, then, yes, you will find the behaviour between the base and emitter terminals to be as a diode. (So it should be, since it's a PN junction.)

But we always use 3 when it's conducting, one's the base, the other "collects" the current and the other "emits" the current...

If he was reading this, he might reassess that 100% mark you recently scored.

Hehe ;)

You draw a sinewave symmetrical about an axis, as usual. But instead of that axis being 0 volts, you label it 12 volts and redraw your 0 volt line further down the page.

Hmm...how do I know how further down to draw that zero?

 

[I like Serena]

Hey Fp!

I am aware of it, but we're not talking about PNP or NPN, we're talking about what is a common emitter, and in this case the emitter must point outside. So common emitter only transpires in NPN...no?

From wikipedia (bolding mine):

"In electronics, a common-emitter amplifier is one of three basic single-stage bipolar-junction-transistor (BJT) amplifier topologies, typically used as a voltage amplifier. In this circuit the base terminal of the transistor serves as the input, the collector is the output, and the emitter is common to both (for example, it may be tied to ground reference or a power supply rail), hence its name. "
Is the emitter (the one with the arrow thingy in it) connected to ground or power?

But we always use 3 when it's conducting, one's the base, the other "collects" the current and the other "emits" the current...

As you saw in your last thread about VBE=±0.7 V, the arrow thingy behaves like a diode.

If you have a voltage drop in the direction of the arrow, the transistor is conducting.
Furthermore that voltage drop is 0.7 V.
The extra side effect is the the connection between collector and emitter is also conducting.

If you have a voltage drop opposite to the direction of the arrow, the transistor is blocking.
Just like a diode. ;)

Hmm...how do I know how further down to draw that zero?

Can you draw a graph of a "direct current whose value is 12V"?
(Note that this is not a sinusoid.)

And can you also draw a graph of a "sinosoidal signal V(t) = 2.2 sin 314t".
What is its amplitude?
(You're allowed to add little drawings of flowers and such. )

Can you finally draw a graph that sums both graphs?
What is its average?

 

[Ratch]

Femme_physics,

No matter what value of R I have, as long as the source voltage (16V in our case) bigger than the zener's diode voltage specification (6V in our case), the zener diode will conduct in this case (despite being backwards biased). Is that a correct assumption to make?

No, it is not. The resistor can be so high that it starves out the diode. That means it reduces the current to its reverse saturation value so it does not enter the reverse breakdown region, and operate as a zener anymore. The zener must have a minimum amount of current to operate as a zener. How much depends on the zener.

I'm asked if this connection is CE (Common Emitter)

The inputs and output have to be designated before the question can be answered.

Given a sinosoidal signal V(t) = 2.2 sin 314t. This signal rides on a direct current whose value is 12V. You may say that the average voltage of the signal is 12√2 V

What is the average of a constant value current/voltage? What is the average of a sinusoidal without any offset? A rectified sine is not sinusoidal anymore. Averages are additive.

Ratch

 

[lazyaditya]

Femme_physics,



No, it is not. The resistor can be so high that it starves out the diode. That means it reduces the current to its reverse saturation value so is does not enter the reverse breakdown region and operate as a zener anymore. The zener must have a minimum of current to operate as a zener. How much depends on the zener.



The inputs and output have to be designed before the question can be answered.



What is the average of a constant value current/voltage? What is the average of a sinusoidal without any offset? A rectified sine is not sinusoidal anymore. Averages are additive.

Ratch

True :)

 

[NascentOxygen]

I am aware of it, but we're not talking about PNP or NPN,

We shouldn't be, but you started it! :tongue2: Your reference to the direction of the arrow somehow being relevant here, when it has no relevance, is what led to this diversion.

we're talking about what is a common emitter, and in this case the emitter must point outside.

Maybe your teacher used a phrase something like "to the outside" and you thought he was talking about the arrow's direction? But he wasn't talking about the arrow's direction; that has no bearing on Common Emitter, or Common Collector, or any configuration. Better refer to your class notes or textbook for the distinguishing feature of a Common Emitter configuration.

So common emitter only transpires in NPN...no?

No. Common Emitter is as valid for PNP as NPN.

But we always use 3 when it's conducting, one's the base, the other "collects" the current and the other "emits" the current...

True. But suppose you needed a small diode, and none were available, you could employ a transistor but leave one of its leads (say, the collector) unconnected, and you'd have yourself a PN junction, AKA a diode.

When relating the BE junction to it's 0.6V drop, it might be better to refer to it as a PN junction (which it is and always will be) rather than a diode (a very strange one when it's involved with transistor action).

Hmm...how do I know how further down to draw that zero?

If your scale is 1 cm = 1 volt, about 12 cm down.

 

[Antiphon]

2) is a common collector amp because the collector is common to input and output.

 

[Femme_physics]

From wikipedia (bolding mine):

"In electronics, a common-emitter amplifier is one of three basic single-stage bipolar-junction-transistor (BJT) amplifier topologies, typically used as a voltage amplifier. In this circuit the base terminal of the transistor serves as the input, the collector is the output, and the emitter is common to both (for example, it may be tied to ground reference or a power supply rail), hence its name. "



Is the emitter (the one with the arrow thingy in it) connected to ground or power?

In question 2 in the original question drawing? Apparently not.

I imagine the emitter is regardless the lowest one (without the arrow in this case), and the collector is always the upper one (with the arrow in this case).

As you saw in your last thread about VBE=±0.7 V, the arrow thingy behaves like a diode.

If you have a voltage drop in the direction of the arrow, the transistor is conducting.
Furthermore that voltage drop is 0.7 V.
The extra side effect is the the connection between collector and emitter is also conducting.

If you have a voltage drop opposite to the direction of the arrow, the transistor is blocking.
Just like a diode. ;)

But it's not really a diode, is it? It's an electrical construct that acts like a diode, I think.

Can you draw a graph of a "direct current whose value is 12V"?
(Note that this is not a sinusoid.)

And can you also draw a graph of a "sinosoidal signal V(t) = 2.2 sin 314t".
What is its amplitude?
(You're allowed to add little drawings of flowers and such. )

Can you finally draw a graph that sums both graphs?
What is its average?

One step at a time :) Is this alright? ->

http://img543.imageshack.us/img543/272/avgm.jpg

(Ahh...forgot flower!)

No, it is not. The resistor can be so high that it starves out the diode. That means it reduces the current to its reverse saturation value so it does not enter the reverse breakdown region, and operate as a zener anymore. The zener must have a minimum amount of current to operate as a zener. How much depends on the zener.

About how high, and how do I see it in calculations?

The inputs and output have to be designated before the question can be answered.

That's a question directly from a test, they weren't designated and I can only answer "true or false" on whether it is CE or not CE.

We shouldn't be, but you started it! Your reference to the direction of the arrow somehow being relevant here, when it has no relevance, is what led to this diversion.

Because how can an emitter on a PNP be common to both? In PNP, it emits current towards the collector.

Maybe your teacher used a phrase something like "to the outside" and you thought he was talking about the arrow's direction? But he wasn't talking about the arrow's direction; that has no bearing on Common Emitter, or Common Collector, or any configuration. Better refer to your class notes or textbook for the distinguishing feature of a Common Emitter configuration.

I don't have class notes about it and we technically finished the material, which I think it complete bollocks since we didn't really go over some important stuff like this, that appeared for instance in last year's test (we're doing an external test not composed by our teacher).

No. Common Emitter is as valid for PNP as NPN.

Well, if you say so... I thought that PNP looks more like common collector since ->

http://img152.imageshack.us/img152/9350/pnpnpn.jpg

True. But suppose you needed a small diode, and none were available, you could employ a transistor but leave one of its leads (say, the collector) unconnected, and you'd have yourself a PN junction, AKA a diode.

I hadn't thought of that. Do people do it a lot, or is it purely for theory?
Seems like using a diode would be cheaper.

If your scale is 1 cm = 1 volt, about 12 cm down.

Alright, I want to see if my idea for a graph is correct so far so I'll wait with it :)

 

[lazyaditya]

One step at a time :) Is this alright? ->

http://img543.imageshack.us/img543/272/avgm.jpg

(Ahh...forgot flower!)



Alright, I want to see if my idea for a graph is correct so far so I'll wait with it :)

why is there a rectified waveform ?

 

[lazyaditya]

And the emitter is always the one with the arrow regardless of its position!

 

[lazyaditya]

Transistor is like two "diodes" working in opposite direction and you when emitter to base junction is reverse biased considering collector to base junction reverse biased it acts in "offset mode" !

But if collector- base juntion is forward biased and emitter base junction is reverse biased it won't work in offset mode and there will be some of the current flowing !

so both the biasings have to be considered !

 

[Femme_physics]

why is there a rectified waveform ?

Because they say it's "DC" yet it originates from AC...so a rectified AC would look like that

And the emitter is always the one with the arrow regardless of its position!

OHH! Are you telling me that ->

http://img269.imageshack.us/img269/8348/elusiveemitter.jpg Really?

 

[lazyaditya]

arrow is always located in the emitter and points towards the direction of conventional current !

 

[lazyaditya]

In your question there was never a "rectifier" word used i guess ? it was only written that ac signal rides over dc signal ?

 

[I like Serena]

In question 2 in the original question drawing? Apparently not.

Yep. In the original question drawing, the emitter is not connected as a common emitter.

I imagine the emitter is regardless the lowest one (without the arrow in this case), and the collector is always the upper one (with the arrow in this case).

Well, the arrow thingy always identifies the emitter, but I think you already got that by now.

But it's not really a diode, is it? It's an electrical construct that acts like a diode, I think.

It's really a diode.
Actually, physically it's 2 diodes lumped together into 1 component.

One step at a time :) Is this alright? ->

(Ahh...forgot flower!)

Not it's not alright - there's no flower in it!

But yes, this is actually the last drawing I asked for.
The only thing wrong with it is what you marked as average.
Did you mix up average with RMS?

Oh, and I see 2 sinusoids in there.
What's the purpose of the 2nd one?

I hadn't thought of that. Do people do it a lot, or is it purely for theory?
Seems like using a diode would be cheaper.

Yeah, a diode is cheaper, but not by much. ;)
They are both very cheap.

Alright, I want to see if my idea for a graph is correct so far so I'll wait with it :)

Well, now you know it needs flowers!

 

[Femme_physics]

Not it's not alright - there's no flower in it!

But yes, this is actually the last drawing I asked for.
The only thing wrong with it is what you marked as average.
Did you mix up average with RMS?

Oh, and I see 2 sinusoids in there.
What's the purpose of the 2nd one?

Perhaps my confusion is at the words "rides on". I never seen it used before. It basically means that the max point of that DC wave is the minimum point of the sinosoindal wave?

Well, now you know it needs flowers!

Heh :)
Did you mean IRL or in paper?

It's really a diode.
Actually, physically it's 2 diodes lumped together into 1 component.

Oh. If there are two diodes there, why are they only showing us one in the schematics?

I would like to repeat this question also:

The resistor can be so high that it starves out the diode. That means it reduces the current to its reverse saturation value so it does not enter the reverse breakdown region, and operate as a zener anymore. The zener must have a minimum amount of current to operate as a zener. How much depends on the zener.

How high, and how do I see it in calculations?

 

[I like Serena]

Perhaps my confusion is at the words "rides on". I never seen it used before. It basically means that the max point of that DC wave is the minimum point of the sinosoindal wave?

Mmmh, it means that the 2 graphs are summed.
It's also called superimposed.

It means that if the sinusoid is zero, the combined graph is 12 V.
And when the sinusoid takes on its maximum value (which one is that?), that the combined graph is 12 V plus this maximum value.

Heh :)
Did you mean IRL or in paper?

Uhh... both?

Oh. If there are two diodes there, why are they only showing us one in the schematics?

Yeah well, they are lumped together in some weird way, so that that the part with an arrow behaves like a regular diode, but the extra leg shows some weird side effects.

Check this out: they use 2 type of materials, called "P" and "N".
Glue them against together, like "P-N", and you have a diode.
Put an extra piece against it, like "N-P-N" or "P-N-P" and you have a transistor.

I would like to repeat this question also:

How high, and how do I see it in calculations?

From wikipedia (bolding mine):

The value of R must satisfy two conditions:

1. R must be small enough that the current through D keeps D in reverse breakdown.
   The value of this current is given in the data sheet for D.
   For example, the common BZX79C5V6[4] device, a 5.6 V 0.5 W zener diode, has a recommended reverse current of 5 mA.
   If insufficient current exists through D, then U_OUT will be unregulated, and less than the nominal breakdown voltage (this differs to voltage regulator tubes where the output voltage will be higher than nominal and could rise as high as U_IN).
   When calculating R, allowance must be made for any current through the external load, not shown in this diagram, connected across U_OUT.
   
2. R must be large enough that the current through D does not destroy the device. If the current through D is I_D, its breakdown voltage V_B and its maximum power dissipation P_MAX, then $I_D V_B < P_{MAX}$.

In other words, the current must be between 2 bounds, given by the data sheet of the diode.
If the current is too low, the zener diode becomes unpredictable.
And if the current is too high, the zener diode is destroyed.
Btw, I suspect that your question is intended for a kind of ideal zener diode, for which this minimum current is not relevant.

 

[Ratch]

Femme_physics,

How high, and how do I see it in calculations?

Calculate the resistor value in the circuit using the diode reverse saturation current.

Ratch

P.S. It would be nice if you wrote the name of who you are addressing at the beginning of your comments like I did in this post.

 

[I like Serena]

Hi Ratch! :)

P.S. It would be nice if you wrote the name of who you are addressing at the beginning of your comments like I did in this post.

If you recognize your quote, you can assume the OP is addressing you.

I usually allow that person (you in this case) some time to respond to a question.
But if no answer is forthcoming (usually when that person is simply not online) and I have some time, I respond instead if I feel like it.

If someone other than the OP writes a post, we should assume that the OP is addressed, since it's the OP's thread.

 

[Ratch]

Serena,

If you recognize your quote, you can assume the OP is addressing you.

Not everyone does quotes all the time. It is nice to know without guessing who is saying what to whom.

If someone other than the OP writes a post, we should assume that the OP is addressed, since it's the OP's thread.

Lots of times the OP drops out of the discussion, and the dialog is between others. Then you have to look at recent posts and try to guess who is saying what to whom. Best in all cases to label and sign your posts.

Ratch

 

[Femme_physics]

Yes I go by the assumption that people recognize their own quotes, but I'm OK with identifying the people I'm replying to more directly...let see how it goes.

@ILS

Mmmh, it means that the 2 graphs are summed.
It's also called superimposed.

It means that if the sinusoid is zero, the combined graph is 12 V.
And when the sinusoid takes on its maximum value (which one is that?), that the combined graph is 12 V plus this maximum value.

I would assume the max value is 12√2, but I'm not sure in anything anymore.

Is lazyaditya's graph accurate? You'll have to excuse me this issue is somewhat confusing to me.

In other words, the current must be between 2 bounds, given by the data sheet of the diode.
If the current is too low, the zener diode becomes unpredictable.
And if the current is too high, the zener diode is destroyed.



Btw, I suspect that your question is intended for a kind of ideal zener diode, for which this minimum current is not relevant.

More than that-- the min current is not even given. But we do know the resistor is only 100 ohms




@ Ratch

Calculate the resistor value in the circuit using the diode reverse saturation current.

But if the resistor is given, why calculate it?

 

[Ratch]

Femme_physics,

But if the resistor is given, why calculate it?

If you are analyzing the circuit, then you need to calculate it. If you are designing the circuit, then you need to calculate it. If you are given the circuit, then you don't have to worry about it.

Ratch

 

[I like Serena]

I would assume the max value is 12√2, but I'm not sure in anything anymore.

Well, we're talking about the signal "a sinusoidal signal V(t) = 2.2 sin 314t".
What is the maximum value of just that signal?

Is lazyaditya's graph accurate? You'll have to excuse me this issue is somewhat confusing to me.

I take it you mean this one?

Yes, that is what happens.
The sinusoidal signal given in (b), which is AC, rides on top of a direct current signal (DC), resulting in the graph given in (a).

More than that-- the min current is not even given. But we do know the resistor is only 100 ohms

Oh?
I thought the question was: "No matter what value of R I have, as long as ...".
Doesn't that mean that the value of R is not given?

 

[NascentOxygen]

Perhaps my confusion is at the words "rides on". I never seen it used before. It basically means that the max point of that DC wave is the minimum point of the sinosoindal wave?

Did you write what you meant to, here?

What you have is basically 12V DC, but it isn't absolutely smooth. There is a small amount of ripple on it, so the DC fluctuates between, for example, 11V and 13V. And that fluctuation is smooth and regular, like a sine wave. (The 11V and 13V levels I mentioned are not precisely the levels in your case in point. And I don't like using the word "DC" like I have here, but it will have to do.)

Oh. If there are two diodes there, why are they only showing us one in the schematics?

Because everyone knows there are two PN junctions in a transistor. In the early days of transistor use, the original symbol had no arrows, and it was necessary to print the letters "C", "E", and "B" alongside the terminals so constructors would know which was which. Then someone had the brilliant idea https://www.physicsforums.com/images/icons/icon3.gif of reducing visual clutter on schematics by doing away with the lettering, and adding an arrow on just one junction, EB, to distingush one junction from the other. Very important to make the distinction, since in circuit construction the junctions are not interchangeable*.

I would like to repeat this question also:

How high, and how do I see it in calculations?

The finer details are not worth troubling yourself with this at your stage; all in good time. We have to get you through this AC riding on DC conundrum for now. https://www.physicsforums.com/images/icons/icon6.gif Nevertheless, for the Zener Diode all is explained in its I vs. V plot here: http://www.electronics-tutorials.ws/diode/diode_7.html

 

[Femme_physics]

Oh?
I thought the question was: "No matter what value of R I have, as long as ...".
Doesn't that mean that the value of R is not given?

In that hypothetical situation-- yes. But in the example test which this question is taken from, it was given (100 ohms).

Because everyone knows there are two PN junctions in a transistor. In the early days of transistor use, the original symbol had no arrows, and it was necessary to print the letters "C", "E", and "B" alongside the terminals so constructors would know which was which. Then someone had the brilliant idea of reducing visual clutter on schematics by doing away with the lettering, and adding an arrow on just one junction, EB, to distingush one junction from the other. Very important to make the distinction, since in circuit construction the junctions are not interchangeable*.

I accept it, and I won't go into the inner workings of the transistor. But let me ask you this... If the arrow here were to be reversed:

http://img543.imageshack.us/img543/4290/50887855.jpg

Would that have been CE?

The finer details are not worth troubling yourself with this at your stage; all in good time. We have to get you through this AC riding on DC conundrum for now. Nevertheless, for the Zener Diode all is explained in its I vs. V plot here: http://www.electronics-tutorials.ws/diode/diode_7.html

Thanks, I'll keep this for general references

Yes, that is what happens.
The sinusoidal signal given in (b), which is AC, rides on top of a direct current signal (DC), resulting in the graph given in (a).

Did you write what you meant to, here?

What you have is basically 12V DC, but it isn't absolutely smooth. There is a small amount of ripple on it, so the DC fluctuates between, for example, 11V and 13V. And that fluctuation is smooth and regular, like a sine wave. (The 11V and 13V levels I mentioned are not precisely the levels in your case in point. And I don't like using the word "DC" like I have here, but it will have to do.)

To be honest I've a bit lost myself. Let me try to refind the string here...

We have a DC wave whose avg (or rms?) is 12 volts...and since the sinusoinal one rides on it it means...

http://img39.imageshack.us/img39/7414/sinedc.jpg

But I'm not sure of anything anymore..

Well, we're talking about the signal "a sinusoidal signal V(t) = 2.2 sin 314t".
What is the maximum value of just that signal?

Are we talking about one of the formulas here? :

http://img822.imageshack.us/img822/31/formulashere.jpg

 

[I like Serena]

I accept it, and I won't go into the inner workings of the transistor. But let me ask you this... If the arrow here were to be reversed:

Would that have been CE?

No, the emitter is still the same leg, and it still isn't connected to a common input-output wire (or ground).

The collector (the one without the arrow thingy) is connected the a common input-output wire.
So if anything, it's "common collector".

To be honest I've a bit lost myself. Let me try to refind the string here...

We have a DC wave whose avg (or rms?) is 12 volts...and since the sinusoinal one rides on it it means...

But I'm not sure of anything anymore..

Riding on it, does not mean the entire graph is put on top.
It means you should "add" the signals.

Are we talking about one of the formulas here?

Yes.
So what's the maximum voltage of the "sinusoidal signal V(t) = 2.2 sin 314t"?
Can you perhaps draw a graph of it?
And what would be its average (not to be confused with the RMS)?

 

[NascentOxygen]

I accept it, and I won't go into the inner workings of the transistor. But let me ask you this... If the arrow here were to be reversed:

Would that have been CE?

It will make NOT A JOT OF DIFFERENCE to the fundamental topology of the circuit, The "top" terminal is still the emitter, regardless. The only major change that use of an NPN will cause is to require that the polarity of the battery (not shown) powering this amplifier be reversed.

 

[Femme_physics]

No, the emitter is still the same leg, and it still isn't connected to a common input-output wire (or ground).

The collector (the one without the arrow thingy) is connected the a common input-output wire.
So if anything, it's "common collector".

But there are no inputs/outputs shown in this circuit. How can you tell that?

Riding on it, does not mean the entire graph is put on top.
It means you should "add" the signals.

Alright, so I'll use the formula (after next quote)..

Yes.
So what's the maximum voltage of the "sinusoidal signal V(t) = 2.2 sin 314t"?
Can you perhaps draw a graph of it?
And what would be its average (not to be confused with the RMS)?

Before I draw the graph...is this the correct calculation?

http://img842.imageshack.us/img842/7929/sinalpha.jpg

It will make NOT A JOT OF DIFFERENCE to the fundamental topology of the circuit, The "top" terminal is still the emitter, regardless. The only major change that use of an NPN will cause is to require that the polarity of the battery (not shown) powering this amplifier be reversed.

Alright. But like I said to ILS the inputs and outputs are not shown in this case...how can you tell whether it's a CE or not?

 

[lazyaditya]

You can see in your circuit that one of the collector terminal is towards base and one of the collector terminal is towards emitter thus is common to both the base and emitter and therefore it is common collector and not common emitter.

 

[Femme_physics]

You can see in your circuit that one of the collector terminal is towards base and one of the collector terminal is towards emitter thus is common to both the base and emitter and therefore it is common collector and not common emitter.

What do you mean by "towards"?