Op-Amp basic problem with a LED diode

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    Diode Led Op-amp
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Homework Help Overview

The discussion revolves around a circuit involving an operational amplifier (Op-Amp) and an LED diode, with a specific focus on the voltage across the LED and the behavior of the Op-Amp under ideal conditions. Participants are analyzing the circuit's configuration and the implications of the Op-Amp's characteristics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the voltage at the Op-Amp's inputs, current pathways, and the implications of using an ideal Op-Amp model. There are attempts to clarify the voltage gain and the relationship between input and output voltages, as well as questions about the current flow in the circuit.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the behavior of the Op-Amp and the interpretation of the circuit, but there is no explicit consensus on all points raised.

Contextual Notes

Participants are navigating the constraints of the ideal Op-Amp model, which states that no current flows into the inputs. There are ongoing questions about the assumptions made regarding voltage and current pathways in the circuit.

  • #61
so I just write that Vout = Saturated?
 
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  • #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?
 
  • #71
The positive supply is always connected at the top of the op-amp, regardless of where the + and - inputs are.

Ah, really? That's kinda funny. Well, in that case

Vout = 7.8 [V]

The negative supply is always connected at the bottom of the op-amp.
[qupte]What is connected at the bottom of the op-amp?[/quote]

Well, V+ (what I defined it as) is connected to the bottom of the op-amp. Therefor -Vout = V+ = -5.526
 
  • #72
Femme_physics said:
Ah, really? That's kinda funny. Well, in that case

Vout = 7.8 [V]

Yes. That is true in case 1.


Femme_physics said:
Well, V+ (what I defined it as) is connected to the bottom of the op-amp. Therefor -Vout = V+ = -5.526

No, that is connected to the + input of the op-amp.
The negative power supply is specified to the right of that.
 
  • #73
I just realized something, I just need to find Vout in case 1 and case 2. Why did I do all those minuses and pluses for?

In both cases the Vout ways exceeds the limit of the op-amp. In fact, shouldn't the op-amp burn if that happens?
 
  • #74
Femme_physics said:
I just realized something, I just need to find Vout in case 1 and case 2. Why did I do all those minuses and pluses for?

Which minuses and pluses?
Femme_physics said:
In both cases the Vout ways exceeds the limit of the op-amp. In fact, shouldn't the op-amp burn if that happens?

No, it just gets saturated and Vout is just either the positive or the negative power supply.
 
  • #75
Which minuses and pluses?

I noticed I was trying to find -Vout and +Vout in the last few posts which doesn't make sense, since there is only 1 Vout in each of the 2 circumstances. I don't know why I did that. Must've gotten confused.

No, it just gets saturated and Vout is just either the positive or the negative power supply.

Gotcha.

No, that is connected to the + input of the op-amp.
The negative power supply is specified to the right of that.

Vout2 = V- = 6.3V

?
 
  • #76
Femme_physics said:
Vout2 = V- = 6.3V

?

Nooooo... there is an Earth symbol there. :rolleyes:
Shouldn't that be the negative power supply?
 
  • #77
Oh!

So, ok, taking the lower leg into consideration it's 7 - 5.526 = 1.474 [V]
 
  • #78
Femme_physics said:
Oh!

So, ok, taking the lower leg into consideration it's 7 - 5.526 = 1.474 [V]

Huh? :confused:

Where did you get 7 V?
 
  • #79
Well, that's V3
 
  • #80
Femme_physics said:
Well, that's V3

But... V3 is not connected to the op-amp.
Before you had already the proper V+ and V-...
 
  • #81
:(

I'm really confused and lost. A bit more clue-ins, please?
 
  • #82
The op-amp has a pair of power connections. This is where the op-amp gets the power to do what it does. The op-amp can only provide an output voltage that lies between the positive power supply and the negative power supply voltages. If you apply inputs that would cause the op-amp to "want" to create an output that is higher than the positive supply voltage, or less than the negative supply voltage, the op-amp can only reach those limits.

What conditions will make the op-amp drive the output to one power supply level or the other? For an ideal op-amp, ANY voltage difference on the inputs will do it!

For this circuit the positive power supply for the op-amp is Vs = 7.8V. The negative supply voltage is zero because the negative supply lead for the op-amp is connected to ground (zero volts). So the output of the op amp is restricted to the range 0 ≤ Vo ≤ 7.8 .

Does that help?
 
  • #83
Are you telling me that Vout in case 2 = 0V ?
 
  • #84
Femme_physics said:
Are you telling me that Vout in case 2 = 0V ?

I suppose I am :smile:
 
  • #85
Ah...because no matter what's my V there, as long as it's in minus, it automatically turns into 0 because the transistor minus leg is ground?

EDIT:

BTW one of my classmates looked at me oddly when I presented this formula:

Vout = (V+ - V-) x 1000000

As though we don't need it to solve the problem. Is it needed, or not?
 
  • #86
Femme_physics said:
Ah...because no matter what's my V there, as long as it's in minus, it automatically turns into 0 because the transistor minus leg is ground?

Well, op-amp rather than transistor (an op-amp actually contains a whole bunch of transistors). The basic rules are: the op-amp output cannot exceed the power supply levels, and any voltage difference between its inputs will drive it to one level or the other.

This applies to ideal op-amps. Real ones come very close to this behavior, too. So much so that one can usually ignore any discrepancies for purposes of analysis.

EDIT: No, you don't really need that formula, but if it helps you to remember which direction the op-amp is going to drive the output by all means keep it handy. Maybe replace the 1000000 with a big G or A or something that won't attract undue remarks :smile:
 
  • #87
Well, op-amp rather than transistor (an op-amp actually contains a whole bunch of transistors).

Oops! Mistyped. meant op-amp :)

The basic rules are: the op-amp output cannot exceed the power supply levels, and any voltage difference between its inputs will drive it to one level or the other.

This applies to ideal op-amps. Real ones come very close to this behavior, too. So much so that one can usually ignore any discrepancies for purposes of analysis.

Ah, of course! I feel so stupid missing all the hints ILS gave me!
EDIT: No, you don't really need that formula, but if it helps you to remember which direction the op-amp is going to drive the output by all means keep it handy. Maybe replace the 1000000 with a big G or A or something that won't attract undue remarks
LOL. Got it :wink:

Thanks! Will scan full solution soon
 
  • #89
Looks fine to me! :smile:
 
  • #90
Terrific :smile: thanks!
 

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