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

#### 123ryoma12

##### Member
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!

#### 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,

$\bg_white \frac{X_{in}}{1.5 k \Omega} = \frac{ V_- - X_{out}}{12 k \Omega}$

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.

$\bg_white Using V_- = V_+ = 0 V , we have that by rearranging X_{out} = -8 X_{in}$

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 $\bg_white V_- = -3 V$

Now for the output to be 16 V, we want $\bg_white V_+ > V_-$

Hence, $\bg_white -8 X_{in} > - 3 V$

$\bg_white 0 > 8 X_{in} - 3 V$

$\bg_white 8 X_{in} < 3 V$

So $\bg_white X_{in} < 0.375$ 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

##### Member
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,

$\bg_white \frac{X_{in}}{1.5 k \Omega} = \frac{ V_- - X_{out}}{12 k \Omega}$

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.

$\bg_white Using V_- = V_+ = 0 V , we have that by rearranging X_{out} = -8 X_{in}$

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 $\bg_white V_- = -3 V$

Now for the output to be 16 V, we want $\bg_white V_+ > V_-$

Hence, $\bg_white -8 X_{in} > - 3 V$

$\bg_white 0 > 8 X_{in} - 3 V$

$\bg_white 8 X_{in} < 3 V$

So $\bg_white X_{in} < 0.375$ 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

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#### 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

##### Member
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".

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