OP Amp Comparators: V+ > V- for +16V Output

In summary, the conversation discusses the solution to a problem involving V+, V-, Vout, and X(in). It is determined that for Vout to be +16V, V+ must be greater than V-, and for it to be -16V, V+ must be less than V-. The conversation also addresses the correct value for Xin in order for Yout to be +16V, which is < 0.375. The graph provided in the conversation is incorrect, as it shows the transition of Yout occurring at -1V instead of -3V.
  • #1
123ryoma12
8
1

Homework Statement


https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/73/73750-4bf3d2aa03a724fcf0b13a6b1111902f.jpg

The Attempt at a Solution


V+ > V- for Vout to be +16V
V+ < V- for Vout to be -16V
V+ = -8x(in)
V- = (-15x18)/30 + (15x12)/30 = -3
My attempt on it was that -8X(in)V greater or equal to -3V for Yout to +16V
(EDIT: Then that would be Xin < 0.375 for it to be +16V
But the answer is saying the opposite.)

But this was the solution and it showed the opposite (that -8x(in) less that or equal to -3)[/B]
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/73/73751-f11fe983c5d64cca6eb97cffb930e594.jpg This was the answer
 

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  • #2
My attempt on it was that -8X(in)V greater or equal to -3V

Do you mean.. Yout is 16V if -8X(in)V greater or equal to -3V ?

In which case that is correct.

But this was the solution and it showed the opposite (that -8x(in) less that or equal to -3)

Their solution looks correct to me. As a check try setting Xin = 1V and work out Xout, Yin, Y+ and Yout.
 
  • #3
My mistake for not completing my answer.
And yes, that's what i meant, When -8(xin) was greater than -3 then the Vout would be +16V
Then that would be Xin < 0.375 for it to be +16V
But the answer is saying the opposite.
 
  • #4
Yout is -16V when Xin < 0.375
 
  • #5
123ryoma12 said:
My mistake for not completing my answer.
And yes, that's what i meant, When -8(xin) was greater than -3 then the Vout would be +16V
Then that would be Xin < 0.375 for it to be +16V
But the answer is saying the opposite.

No the graph shows that when Xin is < 0.375 then Yout is +16v. For example at t=0.
 
  • #6
The output graph is wrong. The transition of Yout going from -16V to +16V occurs when Xout = -3V, not -1V as shown.
 
  • #7
123ryoma12 said:
Yout is -16V when Xin < 0.375
No.
 
  • #8
Sorry guys, I just read the graph incorrectly.
Thanks for pointing that out.
 

FAQ: OP Amp Comparators: V+ > V- for +16V Output

What is an OP Amp Comparator?

An OP Amp Comparator is an electronic device that compares two input voltages and produces an output voltage based on the comparison. It is commonly used in electronic circuits to make decisions based on the voltage levels of the inputs.

How does an OP Amp Comparator work?

An OP Amp Comparator works by amplifying the voltage difference between the two input terminals, known as V+ and V-. If V+ is greater than V-, the output of the OP Amp will be a positive voltage, and if V- is greater than V+, the output will be a negative voltage.

What does V+ > V- mean for +16V Output?

This notation means that the voltage at the V+ input must be greater than the voltage at the V- input in order for the output to be +16V. If the V+ input is less than the V- input, the output will be negative.

What is the purpose of using V+ > V- for +16V Output in an OP Amp Comparator?

The purpose of using this configuration is to create a voltage threshold for the output. By setting V+ to a specific voltage level and V- to a lower voltage level, the comparator can be used to trigger other electronic components in a circuit when the input voltage reaches a certain level.

Are there any limitations to using V+ > V- for +16V Output in an OP Amp Comparator?

Yes, there are some limitations to using this configuration. If the input voltages are too close in value, the output may become unstable and produce unpredictable results. Additionally, the output voltage cannot exceed the power supply voltage, so using +16V as the output limit may not be possible if the power supply voltage is lower.

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