Who's doing it this year? (1 Viewer)

[123ryoma12]

The test is in 3 days. I'm curious on how many people are doing it with me.

 

[anomalousdecay]

I did it 2 years ago and have completed further study in pretty much all of the option topic. Feel free to post up marathon like questions in a thread here and I can help answer!

 

[123ryoma12]

I did it 2 years ago and have completed further study in pretty much all of the option topic. Feel free to post up marathon like questions in a thread here and I can help answer!

Can you please help me with this
https://www.physicsforums.com/threads/op-amps-comparators.839399/#post-5267710

 

[anomalousdecay]

Yep so the flaw with HSC is that they don't teach you Kirchoff's laws. Although you got the answer out in the end, the following method may give you a bit of insight into how this actually occurs. If you don't understand it, then don't worry about the below and just use your old method to find it.

Effectively, the current entering the op amp V- terminal is zero for the ideal case! Now if we think about it, the current entering the node at V- is equal to the current exiting the node (this is Kirchoff's current law). This occurs only for the case of the negative feedback amplifier!

Forming an equation,



Recall that in the ideal inverting configuration, there is infinite input resistance so that the current entering the op amp is zero and so that the voltage at V- and V+ are equal. This is for the inverting op amp configuration and it does not occur in every op amp circuit configuration.



That gives you the that part of the graph.

Now the y output is either saturated at 16 V or -16 V. It is determined by the comparator circuit including op amp Y.


I got the same value as you, such that


Now for the output to be 16 V, we want


Hence,






So for the output to be 16 V.

Now we know that once the V+ input is less than 0.375 V, we have an output of 16 V. Similarly, if we put a larger value for X_in, we will get that X_out < - 3 V, so that V- > V+ and hence the output of Y will be -16 V.

Plotting all these results will give you the answer as given for the question.

Hope that helps!

 

[123ryoma12]

Yep so the flaw with HSC is that they don't teach you Kirchoff's laws. Although you got the answer out in the end, the following method may give you a bit of insight into how this actually occurs. If you don't understand it, then don't worry about the below and just use your old method to find it.

Effectively, the current entering the op amp V- terminal is zero for the ideal case! Now if we think about it, the current entering the node at V- is equal to the current exiting the node (this is Kirchoff's current law). This occurs only for the case of the negative feedback amplifier!

Forming an equation,



Recall that in the ideal inverting configuration, there is infinite input resistance so that the current entering the op amp is zero and so that the voltage at V- and V+ are equal. This is for the inverting op amp configuration and it does not occur in every op amp circuit configuration.



That gives you the that part of the graph.

Now the y output is either saturated at 16 V or -16 V. It is determined by the comparator circuit including op amp Y.


I got the same value as you, such that


Now for the output to be 16 V, we want


Hence,






So for the output to be 16 V.

Now we know that once the V+ input is less than 0.375 V, we have an output of 16 V. Similarly, if we put a larger value for X_in, we will get that X_out < - 3 V, so that V- > V+ and hence the output of Y will be -16 V.

Plotting all these results will give you the answer as given for the question.

Hope that helps!

https://www.physicsforums.com/attachments/capture2-png.90685/
This was BOS answer. When Xin <0.375 Yout is -16V not 16V
Which is why im confused
This was 2014 paper btw
http://www.boardofstudies.nsw.edu.au/hsc_exams/2014/pdf_doc/2014-mg-physics.pdf

 

[anomalousdecay]

You are reading the graph incorrectly. Clearly the graph shows that when X_in < 0.375 V, Y_out = 16 V

When X_in > 0.375 V, Y_out = -16 V

 

[123ryoma12]

omg my bad.. idk how i read it incorrectly. thx
sorry for wasting your time

 

[123ryoma12]

I have another question
Can you please help me with question a part v
I don't really know what they're asking for in this question.
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2012exams/pdf_doc/2012-hsc-exam-physics.pdf

 

[anomalousdecay]

Basically, you want a point where the output is zero, which means that V+ = V- and hence there is an equilibrium that can be said about the inputs.

Now, V- is fixed at 5 V (voltage division). We want to find the point when V+ = 5 V.

To do so, using voltage division you can see that we want the thermistor to be 6 k. Refer to the graph which will tell you the point where the thermistor resistance is 6 k.

Does this help and make sense to you?

 

[123ryoma12]

Basically, you want a point where the output is zero, which means that V+ = V- and hence there is an equilibrium that can be said about the inputs.

Now, V- is fixed at 5 V (voltage division). We want to find the point when V+ = 5 V.

To do so, using voltage division you can see that we want the thermistor to be 6 k. Refer to the graph which will tell you the point where the thermistor resistance is 6 k.

Does this help and make sense to you?

Thanks for this. I understood you're working, I just still have no idea what they mean by "equilibrium temperature". It would be a lot easier if they just asked what the temperature is for Vout = zero. Hopefully they word things differently in this year's exam :L

EDIT: I understand that there is an equilibrium between the inputs but I'm completely clueless about "equilibrium temperature".

 

[anomalousdecay]

Just remember for the future that "equilibrium temperature" refers to the temperature at which an equilibrium occurs at the inputs.

The wording is an example of poor communication. But what can you do? Just try what you can from the question and state an assumption of what you think it means at the top of the question.