How to Calculate Vout Using Comparators?

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To calculate Vout using comparators, the formula Vout = A(V+ - V-) is essential, where A is the gain factor. In this case, if V+ is 1uV and V- is grounded (0V), the output would be Vout = 1,000,000 * 1uV, resulting in 1V. It's important to clarify that the op-amp's negative input (V-) may not always be grounded, as it can be connected to other voltages depending on the circuit design. Understanding the op-amp configuration, including power supply connections, is crucial for accurate calculations. Proper comprehension of these concepts is necessary for effective application in circuit analysis.
Lildon
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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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