Op-Amp basic problem with a LED diode

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SUMMARY

The discussion centers on analyzing a circuit involving an ideal operational amplifier (op-amp) and an LED diode, with a specific focus on voltage calculations and current pathways. The voltage across the LED is established at 1.4 volts, while the op-amp configuration is identified as an inverting amplifier with a voltage gain of 5, derived from the resistor values R2 (10kΩ) and R1 (2kΩ). Participants clarify that no current flows into the op-amp inputs, emphasizing the importance of feedback loops and the relationship between input and output voltages in maintaining equilibrium in the circuit.

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  • #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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