How to Calculate Vout Using Comparators?

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Discussion Overview

The discussion revolves around calculating the output voltage (Vout) using comparators and differential amplifiers. Participants explore the relationship between input voltages and the output, particularly in the context of a homework problem involving an op-amp configuration.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the concept of gains and how to approach the problem, indicating a lack of instruction on the topic.
  • Another participant states that a differential amplifier amplifies the difference between the positive and negative input voltages by a gain factor A, providing the formula Vout = A(V+ - V-).
  • There is a question about the specific values of V+ and V- and their difference, leading to a calculation attempt of Vout based on assumed values.
  • Participants discuss the grounding of V- and whether it is indeed at 0 volts, with one participant seeking clarification on the connections in the op-amp diagram.
  • A detailed explanation of the op-amp symbol and its connections is provided, clarifying the roles of the inputs and power supply.
  • One participant expresses relief at the clarification regarding the op-amp connections, indicating a better understanding of the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the values of V+ and V- or their connections, indicating ongoing confusion and differing interpretations of the op-amp configuration.

Contextual Notes

Some assumptions about the circuit configuration and the grounding of inputs remain unresolved, and there are uncertainties regarding the specific values to be used in calculations.

Lildon
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Homework Statement



problem2.png
I'm just not sure how to do this at all really. My instructor never talked about gains and I can't find anything useful on the web. Any help is appreciated.
 
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A differential amplifier amplifies the difference between the + and - input voltages by some gain factor A.

Vout = A(V+ - V-)
 
so then it would be Vout = 1,000,000 * 1uV?
 
Lildon said:
so then it would be Vout = 1,000,000 * 1uV?

What's the voltage on V+? What's the voltage on V-? What's their difference?
 
V+ is 1uV and the other is grounded right? so 1 - 0 = 1uV * 1,000,000. I have hardly been taught this and I'm supposed to just figure it out. is V- not 0?
 
Lildon said:
V+ is 1uV and the other is grounded right? so 1 - 0 = 1uV * 1,000,000. I have hardly been taught this and I'm supposed to just figure it out. is V- not 0?

Your diagram does not show V- as grounded. What's it connected to?
 
Lildon said:
V+ is 1uV and the other is grounded right?
Nooo.

The op-amp symbol is a triangle, and here it is pointing to the right. That's the output. The vertical side on the left is where the inputs are located. There are two of them, one towards the top of that vertical line, and it's marked + because it is the non-inverting input. The other input is located lower down on that vertical line, and it's marked - because the output will have a component that is the inverse of this.

The op-amp also needs to be powered from a battery or some dc power source, and by convention the connections for the dc source are drawn coming out of the oblique sides of the triangle: emerging from the top is the connection to the + terminal of the battery or dc source, and emerging from the bottom oblique side is the connection to the - terminal of the battery. In some cases, instead of having a negative dc source, the designer returns that connection to ground (0 volts), and that's the case here. The vertical line you see at the lower oblique side of the op-amp symbol is not an input, it's the power supply pin of the op-amp going to ground.
 
Okay that makes more sense. I wasn't sure if he was talking about the positive and negative terminals of the voltage source or if it was the the amplifier. Thanks for explaining it.
 
so now it would be Vout = https://www.physicsforums.com/images/icons/icon5.gif
 
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