Jdean902
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#1
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So... What was that?

Anyone else relate or am I the only person that found that hard? I also ran out of time...
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xiaochen
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(Original post by Jdean902)
So... What was that?

Anyone else relate or am I the only person that found that hard? I also ran out of time...
It was soooo difficult T-T

I managed to put an answer to all the questions but idk...
I didn’t revise push pull amplifier so got the n-ch and p-ch mosfet the wrong way round
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Jdean902
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(Original post by xiaochen)
It was soooo difficult T-T

I managed to put an answer to all the questions but idk...
I didn’t revise push pull amplifier so got the n-ch and p-ch mosfet the wrong way round
I did exactly the same, revised it but still put it the wrong way round !
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xiaochen
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(Original post by Jdean902)
I did exactly the same, revised it but still put it the wrong way round !
Oh well literally everyone in my class got it wrong - but it was only like 2 marks so it shouldn’t be too bad...
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Jdean902
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(Original post by xiaochen)
Oh well literally everyone in my class got it wrong - but it was only like 2 marks so it shouldn’t be too bad...
I completely messed up Question 5 too about the microcontroller and maintaining the current <>.<>
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xiaochen
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(Original post by Jdean902)
I completely messed up Question 5 too about the microcontroller and maintaining the current <>.<>
I was literally on the edge of panicking but I managed to think of some answer before the end :’)

I put for the first part that the DAC outputs 3.5v so 0.5v is across the 1ohm resistor. Then the voltage across 1 ohm is detected by the ADC such that the voltage output of the DAC maintains the 0.5A. As the battery charges, the voltage across the resistor will fall so the DAC output needs to increase.

And then a similar thing for the second part except that voltage is now maintained (initially increased from 4v).

I’m not sure!!
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Jdean902
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(Original post by xiaochen)
I was literally on the edge of panicking but I managed to think of some answer before the end :’)

I put for the first part that the DAC outputs 3.5v so 0.5v is across the 1ohm resistor. Then the voltage across 1 ohm is detected by the ADC such that the voltage output of the DAC maintains the 0.5A. As the battery charges, the voltage across the resistor will fall so the DAC output needs to increase.

And then a similar thing for the second part except that voltage is now maintained (initially increased from 4v).

I’m not sure!!
Sounds about right. i mean it makes sense i threw down an answer for the first part but was out of time for the diagram and the second explination OH WELL whatis done is done
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xiaochen
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(Original post by Jdean902)
Sounds about right. i mean it makes sense i threw down an answer for the first part but was out of time for the diagram and the second explination OH WELL whatis done is done
Yeah it’s it’s not over yet we still have elec5 so good luck!!
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