Is it possible to find Vout in this case ?

[Femme_physics]

http://img32.imageshack.us/img32/69/26879182.jpg



Is it possible to find Vout in this case? Voltage divider doesn't seem to work since it's not two resistor, it's a resistor and a voltage drop.

 

[technician]

FP
I think that diode symbol you have drawn is a zener diode. If so then the Vout is 2V... that is what zener diodes do... they set the voltage to what they are designed
It means the voltage across the 1Ω is 4 volts so the current down the series circuit is 4A

 

[gneill]

That's a zener diode, so it wants to clamp (limit) the voltage across itself at 2V. So right away you can say that Vout will be in the neighborhood of 2V. This is true so long as the supply voltage is greater than the zener voltage, where the diode starts conducting.

For a more precise figure for Vout you would need to have more information about the particular zener diode's characteristics. They tend to exhibit some resistance that increases with current flow. This one will be drawing about 4 amps!

 

[Femme_physics]

Yes, it's a zener diode.

If so then the Vout is 2V... that is what zener diodes do... they set the voltage to what they are designed

REally? I didn't know that.

It means the voltage across the 1Ω is 4 volts so the current down the series circuit is 4A

That makes sense! Thanks.

 

[Femme_physics]

That's a zener diode, so it wants to clamp (limit) the voltage across itself at 2V. So right away you can say that Vout will be in the neighborhood of 2V. This is true so long as the supply voltage is greater than the zener voltage, where the diode starts conducting.

For a more precise figure for Vout you would need to have more information about the particular zener diode's characteristics. They tend to exhibit some resistance that increases with current flow. This one will be drawing about 4 amps!

Excellent. Yes, we're in basic zener diodes exercises, yet, so I don't think it has any special characteristics.

Thanks a lot gneill, techhie :)

 

[technician]

It's a pleasure
You keep us on our toes
Have probably given too much help... just realized this is a homework forum

 

[gneill]

Excellent. Yes, we're in basic zener diodes exercises, yet, so I don't think it has any special characteristics.

Thanks a lot gneill, techhie :)

You're welcome, as always!

If it's early days in zener school , then assume an ideal diode: zero on resistance and a sharp turn-on at its specified voltage.

 

[Femme_physics]

It's a pleasure
You keep us on our toes
Have probably given too much help... just realized this is a homework forum

No, I love your help, keep it up please :)

You're welcome, as always!

If it's early days in zener school , then assume an ideal diode: zero on resistance and a sharp turn-on at its specified voltage.

Gotcha.

Actually, this question I asked comes from this question:

http://img35.imageshack.us/img35/8592/myworkp.png

I mostly now have a problem figuring out if I should treat the section just above the diode zener (the place I marked "e") as a crossroad, or it should really be a little loop like this:

http://img27.imageshack.us/img27/8527/webcam1322984748.png

 

[technician]

This is another weird circuit I would say the diagram should have the loop so that the Zener diode (3V?) is in series with the R above it...but... What is that resistor connected to?
Also I do not understand the 3resistors on the left!
Also...this is another circuit with no feedback to the - input so the output is certain to be +Vs or -Vs.
It will be interesting to hear the other responses that come your way.
Thanks for your kind comment

 

[Femme_physics]

I would be very interested, too. And I'd like to know if I can solve it using KCL and KVL only?

 

[gneill]

I would be very interested, too. And I'd like to know if I can solve it using KCL and KVL only?

I don't think you'll need even Ohm's law for this one! If that odd loop of resistors between the +12V supply and the + input is correct (no missing ground or power connection) then what determines the voltage on the + input of the op-amp? Hint: What current flows into op-amp inputs?

 

[Femme_physics]

My apologies!

THIS is the circuit:
http://img190.imageshack.us/img190/6100/oopsgq.jpg

Sorry! Missed a ground spot!

That...changes everything, yes?

 

[gneill]

My apologies!

THIS is the circuit:
http://img190.imageshack.us/img190/6100/oopsgq.jpg

Sorry! Missed a ground spot!

That...changes everything, yes?

Actually, not much changes in terms of the result at the output. Can you determine the voltages present on the op-amp inputs?

 

[technician]

Looks "easy" now !

 

[Femme_physics]

Well, I still wound up in the same dilemma. Is the crossroad I marked "e" looks like really this?

http://img27.imageshack.us/img27/8527/webcam1322984748.png i.e. the crossing wires don't touch each other?

 

[technician]

That is what that symbol means. If they were in contact it would (should !) be shown clearly with a dot on the connection

 

[Femme_physics]

That is what that symbol means. If they were in contact it would (should !) be shown clearly with a dot on the connection

Ah, thanks! So in that case, answering the questions...

A) Find Vout

B) Find Vout when C is shortened to the ground

http://img585.imageshack.us/img585/4126/zelat.jpg

 

[I like Serena]

For (A) the plus supply voltage is not -7.5 V.

After that we'll get to (B).

 

[Femme_physics]

What did I do wrong?

 

[I like Serena]

You picked the wrong (the negative) power supply.

 

[Femme_physics]

Oh, my bad. I think I got it worked out now.

http://img15.imageshack.us/img15/3121/zener1k.jpg

 

[I like Serena]

Much better!
And nice "m" in your [mA]! One thing though.
You write Vout=(V+ - V-)=1
But this is not correct.

You wrote before that V+ was greater than V-, so Vout was the positive voltage supply.
That was correct.

You can also write it like this, but then you should write something like:
Vout=(V+ - V-) 1000000 = 1000000 [V] which is greater than Vss, so Vout=Vss=8.4 [V].

 

[Femme_physics]

Got it fixed. I'll scan my solution when I get home

Much obliged!

 

[Femme_physics]

http://img832.imageshack.us/img832/1952/isnow.jpg

 

[I like Serena]

There!