Understanding Zener Diode Operation

[NascentOxygen]

the inputs and outputs are not shown in this case...how can you tell whether it's a CE or not?

Ah! Now you're asking the right question, FP.

There are conventions in electronics, many of which you've been exposed to throughout your studies, whether you realize it or not. Inputs, by convention, are drawn to the left. Outputs to the right. The ground bus is horizontal and drawn lower down on the page than the [single] supply rail. Further, because the question asks about Common Emitter, you can infer that the schematic represents the bare bones of an amplifer. It is a "simplified" drawing because it doesn't show incidental detail such as bias arrangements, power supplies, and coupling. It shows just the basic topology, and that's all that is required to answer the question. Assuming this isn't a lightning arrestor, then something logical and obvious must be associated with what are pictorially shown as "free" ends of the resistors.

Perhaps this was a challenge question? Or maybe the aim was to ensure there'd be no further embarrassment to the examiner of someone scoring another 100%.

 

[I like Serena]

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

It has the standard H-form (an H on its side) of a circuit with an input and an output.
2 wires on the left (for input) and 2 wires on the right (for output), with the bottom 2 connected to each other.
Often the bottom wire is connected to earth.

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

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

No. I'm afraid you're making this more complex than it is.

Where did you get U(diac)=12 V?
I don't see a 12 V in "V(t)=2.2 sin 314t".

 

[Femme_physics]

You'll have to excuse me, I was a bit worn out.

I'm back now, and I asked my teacher about it.

He said the average voltage is just 12V! Since we only look at the DC voltage number since the average of a sine wave is its middle anyway (or zero).

Can I get a confirmation?

 

[lazyaditya]

Yes !

 

[Femme_physics]

Ah...thank you :)

Things are picking up!

 

[I like Serena]

He said the average voltage is just 12V! Since we only look at the DC voltage number since the average of a sine wave is its middle anyway (or zero).

Can I get a confirmation?

Ah, you're cheating. ;p

But... why is it 12 V?
And what's that sine riding on top of a direct current?

 

[Femme_physics]

Hehe

Well, cheating or not, I just got really frustrated (not blaming u guys I just felt I had a brain failure at understanding terms and what they want from me). So, I had to ask for some clarity from my teacher who explained me how this whole thing works.

That sine is just a sinosoidal wave, riding over a 12 volts DC (straight line in a graph). 12V is the average voltage of that wave, because the positive and the negative of he sine wave negate each other.

 

[I like Serena]

Good. ;)
So what's the amplitude of that sinusoidal wave?

(And know that his omnipotence is watching.)

 

[Femme_physics]

That would be just 2.2 x 2 = 4.4 volts since the amplitude of a sine wave is 2Umax.

 

[I like Serena]

There you go! :)

Note that you gave the so called "peak-to-peak amplitude" which is indeed 2Umax.
The "amplitude" is just Umax.

 

[Femme_physics]

Oh, so the amplitude is already given as 2.2Volts, yes?

 

[I like Serena]

Oh, so the amplitude is already given as 2.2Volts, yes?

Yes. :)

 

[Femme_physics]

Many thanks dungeon master :)

 

[I like Serena]

You're welcome, my happy dwarf amazon. ;)