Op-Amp basic problem with a LED diode

In summary, the voltage at the -input is +2V (based on the minus leg), so current through R1 is 1mA. The voltage at the output is 10V + 2V (based on the battery), so current through the LED is 5.3mA.
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  • #37
Femme_physics said:
I'll just fix the mA later... :) forgot

Now if you'd only add Vout to your drawing (and preferably Vin too), you're set!
(Since Vout still drops from out of the air now.)

Edit: You might even mention it's a non-inverting amplifier. That would make a grand impression!

Oh, and while you're at it, perhaps you can prettify my little milli...
 
  • #39
Now it truly looks like micro (μ or u)!
Do you want to make me cry? :cry:
And apparently you forgot the second time.

Oh, and I just noticed.
At the top you write "RL = ?".
But you already know RL. I think you meant "ILS = ?".
 
  • #40
I like Serena said:
Now it truly looks like micro (μ or u)!
Do you want to make me cry? :cry:
And apparently you forgot the second time.

Oh, and I just noticed.
At the top you write "RL = ?".
But you already know RL. I think you meant "ILS = ?".

Ah, ok ok! Got it now :wink:

http://img580.imageshack.us/img580/8129/finalfixe.jpg
 
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  • #41
Femme_physics said:
Ah, ok ok! Got it now :wink:

Yep! That's it! Don't change anything! :smile:
 
  • #43
Femme_physics said:
Good :smile: :wink:

Thanks a bunch!

So now I could move on to my next exercise...think I'll keep it with this thread. Does it all look right? I got to find Vout when the switch is in state 1 and when it's in state 2

http://img36.imageshack.us/img36/1464/blackblacke.jpg

Not quite.
First, I get a different value for V+ (only by a little).
And I believe it's not in mV, but in V.
(Nice "m" by the way! :smile:)

Then you need to know that the real response of an op-amp is given by:
Vout = (V+ - V-) AOL
where AOL is literally "a zillion", but if you want, you can substitute "a million".
You should substitute this in your formula.

There is one catch.
If Vout becomes to high or too low, the op-amp is "saturated".
So Vout can never become higher that the positive voltage supply.
And Vout can never become lower than the negative voltage supply.

In practice this means that if V+ is higher than V-, then Vout is the positive power supply.
And if V+ is lower than V-, then Vout is the negative power supply.
 
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  • #44
Btw, how is it that you still have access to the scanner?
 
  • #45
Btw, how is it that you still have access to the scanner?

Still? I'm at home. Dad's napping. I ain't bothering him. :)
Not quite.
First, I get a different value for V+ (only by a little).
And I believe it's not in mV, but in V.
(Nice "m" by the way! )

hehe. had to do a last minute fix-up!

Well, you're right, I got a different result too. V = 5.5263V

I guess I must've mistyped something in the calculator the first time. I should've known that result doesn't make sense.

Is that the only mistake?

There is one catch.
If Vout becomes to high or too low, the op-amp is "saturated".
So Vout can never become higher that the positive voltage supply.
And Vout can never become lower than the negative voltage supply.

Oh yea, I remember that!

In practice this means that if V+ is higher than V-, then Vout is the positive power supply.
And if V+ is lower than V-, then Vout is the negative power supply.

Got it! :smile:
 
  • #46
Femme_physics said:
Still? I'm at home. Dad's napping. I ain't bothering him. :)

Okay, I just remember that at other times you could not get to the scanner any more...



Femme_physics said:
hehe. had to do a last minute fix-up!

Well, you're right, I got a different result too. V = 5.5263V

I guess I must've mistyped something in the calculator the first time. I should've known that result doesn't make sense.

Is that the only mistake?

That's the only calculation error.
But your answers are wrong due to my other remark.


Femme_physics said:
Oh yea, I remember that!

Got it! :smile:

Got it? Then what's your conclusion?
 
  • #47
Okay, I just remember that at other times you could not get to the scanner any more...

Hmm...yea you're right. Oh! That's because we moved the scanner and stationary PC to the livingroom, so I'm not bothering anyone here much. Heh, yea, I forgot! I couldn't use the scanner back then... damn, good old days :smile:

That's the only calculation error.
But your answers are wrong due to my other remark.

But my Vout results don't exceed the 7.8V I see signified on the op-amp.
 
  • #48
Femme_physics said:
Hmm...yea you're right. Oh! That's because we moved the scanner and stationary PC to the livingroom, so I'm not bothering anyone here much. Heh, yea, I forgot! I couldn't use the scanner back then... damn, good old days

I've got a couple of good memories and pictures from those days. ;)


Femme_physics said:
But my Vout results don't exceed the 7.8V I see signified on the op-amp.

That's because you did not calculate Vout properly.

Try: [itex]V_{out} = (V_+ - V_-) \times 1000000[/itex].
 
  • #49
I don't have have that formula you just gave me on my chart.

Hmm, maybe you can help me with something before I continue to solve it. I want to make a proper list of the op-amp formulas I need. Do you think that's enough?

http://img27.imageshack.us/img27/9494/forform.jpg
 
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  • #50
Here's the corresponding wiki page that I recommend:
http://en.wikipedia.org/wiki/Operational_amplifier_applications
It has really nice pictures, a nice overview, and pretty formulas.

I recommend adding the Comparator (the first one, which you would use for the problem at hand).
You probably won't need the others.

And the formula I gave, is the one formula that exactly describes the behavior of an op-amp (up to saturation).
All the other formulas can be derived from that one.

Btw, this is the first time that we've been up so late! ;)
Any particular occasion?
 
  • #51
Here's the corresponding wiki page that I recommend:
http://en.wikipedia.org/wiki/Operati...r_applications
It has really nice pictures, a nice overview, and pretty formulas.

Great link! will do. Appears to have everything I need.

I recommend adding the Comparator (the first one, which you would use for the problem at hand).
You probably won't need the others.

Will do as well.
And the formula I gave, is the one formula that exactly describes the behavior of an op-amp (up to saturation).
All the other formulas can be derived from that one.
You mean this?[itex]V_{out} = (V_+ - V_-) \times 1000000[/itex]
 
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  • #52
Femme_physics said:
You mean this?


[itex]V_{out} = (V_+ - V_-) \times 1000000[/itex]

Yes.

So... no particular occasion?
 
  • #53
Oh, sorry, I missed the question :smile: No, just got into a solving shvoong! Can't stop! Should I?
 
  • #54
Femme_physics said:
Oh, sorry, I missed the question :smile: No, just got into a solving shvoong! Can't stop! Should I?

Certainly not!
I like it!
Anyway, I have an hour on you and I don't have to work tomorrow.
(Don't you?)
 
  • #55
I do actually, and starting to feel kinda tired. Will continue this tomorrow, hopefully. Thank you Klaas!

And I love it!

G'night for now :)
 
  • #56
Femme_physics said:
I do actually, and starting to feel kinda tired. Will continue this tomorrow, hopefully. Thank you Klaas!

And I love it!

G'night for now :)

Oh well, good night then Or!
 
  • #58
Femme_physics said:
So, I think I finally got it :)

Yep. That's it... only one problem... the power supply of the op-amp can not yield so many volts. ;)


Femme_physics said:
EDIT: In the wiki article you linked me, there's no reference for the formula of the voltage gain of a non-inverter op-amp. Should we add it?

No reference? Isn't it already there in the 3rd circuit application (Non-inverting amplifier)?

Yeah, otherwise we should add it!
 
  • #59
Yep. That's it... only one problem... the power supply of the op-amp can not yield so many volts. ;)

Yea, I didn't thought that made much sense...otherwise one amplifier and I could run this entire world!

So, what did I do wrong?

No reference? Isn't it already there in the 3rd circuit application (Non-inverting amplifier)?

Yeah, otherwise we should add it!

Think we should too! But I already added it to my own notes and that's what matters :) thanks.
 
  • #60
Femme_physics said:
Yea, I didn't thought that made much sense...otherwise one amplifier and I could run this entire world!

So, what did I do wrong?

In the diagram you can see that the op-amp has a positive and negative power supply.
Vout is capped between those 2.
The op-amp is saturated then.
 
  • #61
so I just write that Vout = Saturated?
 
  • #62
Femme_physics said:
so I just write that Vout = Saturated?

Not quite.
What is the positive power supply of the op-amp?
 
  • #63
I like Serena said:
Not quite.
What is the positive power supply of the op-amp?
Ah so 7.8 [V] :) !

7.8 [V] for the positive
-7.8 [V] for the negativeThose are my Vouts

The name of this circuit/op-amp is "comparator" right?
 
  • #64
Femme_physics said:
Ah so 7.8 [V] :) !

7.8 [V] for the positive
-7.8 [V] for the negativeThose are my Vouts

The name of this circuit/op-amp is "comparator" right?
Almost!

What is the negative power supply?
(It is not -7.8 V.)

And yes, this is a "comparator". Good!
 
  • #65
I updated and edited the post above u :)
 
  • #66
Femme_physics said:
I updated and edited the post above u :)

All right. I also edited the post above u :)
 
  • #67
it's not -7.8? but we hit saturation point. isn't that the max lowest voltage?
 
  • #68
Femme_physics said:
it's not -7.8? but we hit saturation point. isn't that the max lowest voltage?

Look at your diagram. What is connected to the negative power supply of the op-amp?
 
  • #69
It's actually the minus side that has the Vs connected to it, so I say -Vout = -7.8 V

And Vout = V+
 
  • #70
The positive supply is always connected at the top of the op-amp, regardless of where the + and - inputs are.
The negative supply is always connected at the bottom of the op-amp.

What is connected at the bottom of the op-amp?
 

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