Is it possible to find Vout in this case ?

  • Thread starter Femme_physics
  • Start date
In summary: Yes, that looks like the correct layout. The crossroad should be marked as "e" and the loop of resistors should be connecting the +12V supply to the + input of the op-amp.The loop of resistors should be connecting the +12V supply to the + input of the op-amp.
  • #1
Femme_physics
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  • #2
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
 
  • #3
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!
 
  • #4
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.
 
  • #5
gneill said:
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 :)
 
  • #6
It's a pleasure:smile:
You keep us on our toeso:)
Have probably given too much help... just realized this is a homework forum
 
  • #7
Femme_physics said:
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 :smile: , then assume an ideal diode: zero on resistance and a sharp turn-on at its specified voltage.
 
  • #8
technician said:
It's a pleasure:smile:
You keep us on our toeso:)
Have probably given too much help... just realized this is a homework forum

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

gneill said:
You're welcome, as always!

If it's early days in zener school :smile: , 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 [Broken]

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 [Broken]
 
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  • #9
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
 
  • #10
I would be very interested, too. And I'd like to know if I can solve it using KCL and KVL only?
 
  • #11
Femme_physics said:
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?
 
  • #13
Femme_physics said:
My apologies!

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

Sorry! Missed a ground spot! :frown:

That...changes everything, yes? :smile:

Actually, not much changes in terms of the result at the output. Can you determine the voltages present on the op-amp inputs?
 
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  • #14
Looks "easy" now !
 
  • #16
That is what that symbol means. If they were in contact it would (should !) be shown clearly with a dot on the connection
 
  • #17
technician said:
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! :smile: 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 [Broken]
 
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  • #18
For (A) the plus supply voltage is not -7.5 V.

After that we'll get to (B).
 
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  • #19
What did I do wrong?
 
  • #20
You picked the wrong (the negative) power supply.
 
  • #22
Much better!
And nice "m" in your [mA]! :smile: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].
 
  • #23
Got it fixed. I'll scan my solution when I get home :smile:

Much obliged!
 
  • #25
There! :smile:
 

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1. Can Vout be calculated using the given information?

It depends on the specific case and the information provided. In some cases, Vout can be easily calculated using basic circuit analysis techniques, while in other cases, more complex equations or simulations may be required.

2. How do I determine Vout in a circuit?

To determine Vout, you will need to analyze the circuit using Kirchhoff's laws and other circuit analysis techniques. This involves identifying all the components in the circuit, calculating the voltage drops and currents at each component, and applying the appropriate equations to find Vout.

3. Is Vout the same as the output voltage?

Yes, Vout is another term for output voltage. It refers to the voltage that is produced by the circuit as a result of the input voltage and the components in the circuit.

4. What factors can affect the value of Vout?

The value of Vout can be affected by several factors, including the values of the components in the circuit, the input voltage, and the circuit topology. Changes in any of these factors can result in a different Vout value.

5. Can Vout be negative?

Yes, Vout can be negative in some cases. This can happen when the input voltage is negative or when the circuit includes components that produce a negative voltage. However, in most cases, Vout is a positive value.

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