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April 29, 2024, 06:43:49 pm

Author Topic: 2009 Insight Q6  (Read 567 times)  Share 

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Andiio

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2009 Insight Q6
« on: June 12, 2011, 02:23:36 pm »
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Why are we supposed to calculate the Vout by multiplying Vgain with Vin when all the past questions I've done has just been reading off the graph? :\

I understand that when you do read off the graph the Vout values are too large and do not correspond with the Vout scale given... but?
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onur369

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Re: 2009 Insight Q6
« Reply #1 on: June 12, 2011, 02:30:52 pm »
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The peaks need calculating to fit in the 'linear' region as what most trial exams try emphasizing.
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Andiio

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Re: 2009 Insight Q6
« Reply #2 on: June 12, 2011, 02:33:56 pm »
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The peaks need calculating to fit in the 'linear' region as what most trial exams try emphasizing.

So, pretty much if the question asks you to 'estimate' the value of Vout or if, after reading the graph, the values of Vout do not fit within the linear/operating region then we're simply meant to calculate Vout?
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onur369

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Re: 2009 Insight Q6
« Reply #3 on: June 12, 2011, 02:38:15 pm »
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Not really, but remember this is VCAA were talking about. Do your working out like your trying to explain or demonstrate to a kid. Markers mark hundreds of them and they wont be bothered looking deeply into your working out, better be safe than sorry.

Gain= Vout/Vin

Vin is giving thus find gain from original graph, dont worry about negativeness of gradient than do 1.5 x Gain

Gain is about 2500, 2500x1.5 = 3.75V
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onur369

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Re: 2009 Insight Q6
« Reply #4 on: June 12, 2011, 02:39:26 pm »
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oops that 1.5 should be 1.5x10^-3 as graph gives it as mV.
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