Electronic circuit with Op-Amp, Switch and Diodes

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Discussion Overview

This thread discusses an electronic circuit involving operational amplifiers (op-amps), switches, diodes, and lightbulbs. Participants explore the circuit's functionality, analyze voltage and current calculations under different switch conditions, and clarify the behavior of the op-amp in the circuit.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants identify the op-amp configuration as a non-inverting op-amp, while others argue it is an inverting op-amp, leading to confusion about the output voltage signs.
  • There is a discussion about the output voltage (Vout) being -8V when the switch is in position 1 and +8V when in position 2, although some participants question the accuracy of these values.
  • Participants express uncertainty about the voltage at point C, with claims that it will not be zero, contradicting earlier assertions.
  • There are differing views on how the circuit operates based on the switch position, with some participants explaining the current flow and others questioning the implications of the battery's orientation.
  • One participant suggests that the action of the circuit should be explained in terms of inputs and outputs, while another expresses frustration about how to approach the open-ended question regarding circuit action.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the op-amp configuration or the output voltage values, indicating multiple competing views and unresolved questions regarding the circuit's behavior.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the circuit's operation, particularly concerning the voltage calculations and the interpretation of the op-amp's behavior in different switch positions.

Who May Find This Useful

This discussion may be useful for students studying electronic circuits, particularly those working with op-amps and seeking to understand circuit behavior under varying conditions.

  • #31
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  • #32
No. Both wrong!

You fixed some errors, but introduced more. You keep jumping ahead before your error is sorted out.

Look what you've done to the inputs: you have current going the wrong way from the battery. Current should enter the external circuit from the long side (+) of the cell symbol.
 
  • #33
  • #34
Let's go back to the sketches here, and just focus on the top one https://www.physicsforums.com/showpos...6&postcount=22

No need for any more sketches. You correctly figured out which diode is conducting. So you know that the circuit is operating, you just need to account for why it does.


Because there's voltage?

Look what you've done to the inputs: you have current going the wrong way from the battery
That's because it's an inverter
 
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  • #35
Femme_physics said:
Because there's voltage?

That's because it's an inverter

Uhh :rolleyes:?

Current flows from + to -.
So it (almost) always flows away from the plus pole of a battery (the long line) and it (almost) always flows toward the minus pole of a battery (short line).
Certainly in this case it does.
However, in some of your drawings it does, but in others it doesn't...
 
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  • #37
http://img535.imageshack.us/img535/5793/t2774.gif

Tiny discordant note is your comment "L1 lights more brightly". Why?
 
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  • #38
Because it has more current going through it?
 
  • #39
Femme_physics said:
Because it has more current going through it?
Hmmm <checking back sees they have different resistances>. Quite right! :wink:

Well, almost certainly right. Providing they have equal voltage ratings, the higher current one should be brighter.
 
  • #40
Femme_physics said:
Ok, I think I got it... :)

Yep. :)
 

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