Electronic circuit with Op-Amp, Switch and Diodes

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SUMMARY

The discussion centers on an electronic circuit involving Operational Amplifiers (Op-Amps), a double-pole double-throw (DPDT) switch, diodes, and lightbulbs. The circuit operates with two configurations based on the switch position, affecting the output voltage (Vout) and the illumination of the lightbulbs. When the switch is in position 1, the output voltage is -8V, lighting up the red lightbulb (L1), while in position 2, the output voltage is +8V, illuminating the green lightbulb (L2). The Op-Amp configuration used is an inverting configuration, and the calculations for voltages at points a, b, and c are critical for understanding the circuit's behavior.

PREREQUISITES
  • Understanding of Operational Amplifier (Op-Amp) configurations, specifically inverting and non-inverting.
  • Familiarity with circuit components such as diodes and lightbulbs, including their electrical characteristics.
  • Knowledge of Kirchhoff's Voltage Law (KVL) for circuit analysis.
  • Ability to perform basic electrical calculations, including Ohm's Law and voltage/current relationships.
NEXT STEPS
  • Study the principles of inverting and non-inverting Op-Amp configurations in detail.
  • Learn how to analyze circuits using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL).
  • Explore the characteristics and applications of diodes in electronic circuits.
  • Investigate the effects of varying resistance on current flow in series and parallel circuits.
USEFUL FOR

Electronics students, hobbyists working with Op-Amps, and engineers designing circuits that involve switching mechanisms and light indicators.

  • #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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