Drawing the DC Model of a Circuit

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SUMMARY

This discussion focuses on drawing the DC model of a circuit, specifically addressing the treatment of operational amplifiers (OP-Amps) and capacitors in DC analysis. It is established that all capacitors, including C1, C2, and C3, act as open circuits at DC, and the leftmost OP-Amp, functioning as a voltage follower, can be simplified by connecting pin 3 directly to pin 5. The second OP-Amp's output (pin 6) remains relevant despite being connected to capacitor C2, which is grounded, and R1 must be retained in the circuit analysis while R3 can be disregarded due to its connection to ground.

PREREQUISITES
  • Understanding of operational amplifiers and their configurations
  • Knowledge of DC circuit analysis principles
  • Familiarity with capacitor behavior in DC circuits
  • Basic circuit simplification techniques
NEXT STEPS
  • Study the role of voltage followers in operational amplifier circuits
  • Learn about the impact of capacitors in AC versus DC analysis
  • Explore circuit simplification methods for complex networks
  • Investigate the implications of grounding components in circuit design
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Electrical engineering students, circuit designers, and anyone involved in analyzing or designing DC circuits will benefit from this discussion.

KTiaam
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Homework Statement



[Picture of Problem](http://i.imgur.com/7q4yj8O.png)

Homework Equations


[/B]
* It is dealing with drawing the DC model of a circuit
* Now i know when drawing a DC Model, all capacitors are open (or ground) and AC sources are grounded as well.
* In this specific problem, i know that the left most OP-Amp is a Voltage follower, meaning that i can just disregard that OP-Amp, and connect a wire form pin 3 to pin 5 and treat it as just a wire connecting those two pins.
* The second OP-Amp on the right is going me trouble, however. Pin 6 goes straight into a capacitor (C2) which is ground. So does that mean that the OP-Amp does not exist?
* Also from my understanding R3 does not exist, because both ends of the resistor are connected to ground. Also if my logic is right about R3, then R1 does not exist.
* At this point any help is appreciated.
 
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At DC capacitor C1, C2, C3 each represents an open circuit.
 
KTiaam said:
* In this specific problem, i know that the left most OP-Amp is a Voltage follower, meaning that i can just disregard that OP-Amp, and connect a wire form pin 3 to pin 5 and treat it as just a wire connecting those two pins.
Not really. An op-amp as a voltage follower also provides isolation. Pin 3 should never "feel" anything that happens to pin 6 (I think you meant pin 6, the output of the op-amp). And pin 6 should not be able to "see" the impedance at pin 3. You want to just declare the voltage at pin 6 to be whatever the voltage is at pin 3.
* The second OP-Amp on the right is going me trouble, however. Pin 6 goes straight into a capacitor (C2) which is ground. So does that mean that the OP-Amp does not exist?
Not at all. The capacitor is an open circuit at DC.
* Also from my understanding R3 does not exist, because both ends of the resistor are connected to ground. Also if my logic is right about R3, then R1 does not exist.
C2 and R3 effectively don't exist at DC. But R1 most certainly does.
 

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