Op-Amp basic problem with a LED diode

[Femme_physics]

Ah, okay.
I was also wondering what the question of the problem was, since it was missing from your OP.

Ah, was it? Heh, my bad...

Which resistor is RL?
In your diagram you have 2 of them, one of 2 kOhm, and one of 1 kOhm.

Well, I've made a mistake in my last sketch drawing 2 resistors called RL. One of them shouldn't be there. Here's the corrected version with me pointing out the RL.

http://img839.imageshack.us/img839/4700/therez.jpg

----------

FULL correction

http://img194.imageshack.us/img194/3141/dd750full.jpg

And if I read your IL correctly, it is in micro-ampere.
Is that intended?

That's mili-ampere.

 

[I like Serena]

Ah, was it? Heh, my bad...



Well, I've made a mistake in my last sketch drawing 2 resistors called RL. One of them shouldn't be there. Here's the corrected version with me pointing out the RL.

http://img839.imageshack.us/img839/4700/therez.jpg

Oh there it is (x)!

FULL correction

http://img194.imageshack.us/img194/3141/dd750full.jpg

That's mili-ampere.

Looks good now, except for the micro-ampere, or if that is an "M", the mega-ampere.
An "m" should have round caps!
Just like a pi should have nice legs when you use it!

 

[I like Serena]

Oh, and wait!

You used Vin, Vout, Rin, and Rf in your calculations, but they are not in your drawing!
Nor do you explain why and how you use them.

 

[Femme_physics]

Looks good now, except for the micro-ampere, or if that is an "M", the mega-ampere.
An "m" should have round caps!
Just like a pi should have nice legs when you use it!

Well according to this it's just mA

http://www.convertunits.com/info/mili+ampere

Oh, and wait!

You used Vin, Vout, Rin, and Rf in your calculations, but they are not in your drawing!
Nor do you explain why and how you use them.

Vin = V1
Rf = R2
Rin = R1
Vout = well it doesn't really equal anything. it's just Vout :P

 

[I like Serena]

Well according to this it's just mA

http://www.convertunits.com/info/mili+ampere

Well it looks like either μA or MA, but is should be mA.
Notice how round it looks (also in your link)?

Vin = V1
Rf = R2
Rin = R1
Vout = well it doesn't really equal anything. it's just Vout :P

I know! But you should put something like that in your calculation.

 

[Femme_physics]

I'll just fix the mA later... :) forgot http://img17.imageshack.us/img17/4534/800rgy.jpg

 

[I like Serena]

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...

 

[Femme_physics]

There?

http://img818.imageshack.us/img818/8615/upped.jpg

 

[I like Serena]

Now it truly looks like micro (μ or u)!
Do you want to make me 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 = ?".

 

[Femme_physics]

Now it truly looks like micro (μ or u)!
Do you want to make me 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

http://img580.imageshack.us/img580/8129/finalfixe.jpg

 

[I like Serena]

Ah, ok ok! Got it now

Yep! That's it! Don't change anything!

 

[Femme_physics]

Good

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://img217.imageshack.us/img217/6305/thisagain.jpg

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

 

[I like Serena]

Good

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! )

Then you need to know that the real response of an op-amp is given by:
V_out = (V₊ - V_-) A_OL
where A_OL 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.

 

[I like Serena]

Btw, how is it that you still have access to the scanner?

 

[Femme_physics]

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!

 

[I like Serena]

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...

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.

Oh yea, I remember that!

Got it!

Got it? Then what's your conclusion?

 

[Femme_physics]

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

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.

 

[I like Serena]

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. ;)

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

That's because you did not calculate V_out properly.

Try: V_{out} = (V_+ - V_-) \times 1000000.

 

[Femme_physics]

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

 

[I like Serena]

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?

 

[Femme_physics]

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?V_{out} = (V_+ - V_-) \times 1000000

 

[I like Serena]

You mean this?


V_{out} = (V_+ - V_-) \times 1000000

Yes.

So... no particular occasion?

 

[Femme_physics]

Oh, sorry, I missed the question No, just got into a solving shvoong! Can't stop! Should I?

 

[I like Serena]

Oh, sorry, I missed the question 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?)

 

[Femme_physics]

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 :)

 

[I like Serena]

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!

 

[Femme_physics]

So, I think I finally got it :)

http://img546.imageshack.us/img546/9986/finals1really.jpg

http://img404.imageshack.us/img404/2643/finals2really.jpg

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?

 

[I like Serena]

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. ;)

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!

 

[Femme_physics]

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.

 

[I like Serena]

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.