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AQA AS Level Electronics Unit 1 (ELEC1) 14 May 2013 Watch

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    Oh I see (10^2)/16 = 6.25W - As for the op-amp, the maximum output voltage is roughly 2V less that the voltage rails. So if the voltage rails are 12V and 0V, the output of the op-amp will be 10V when high and 2V when low - this is why (like in our paper) sometimes whatever is connected to the output (e.g. an LED) will be on even if the output is low. Though I don't understand how that would fit into our calculation...? Perhaps the report is referring to if you take into account the voltage drop and therefore use 8V in the calculation instead of 10V, in which case you would get 4W which is less than 5W - that would mean the amplifier wouldn't be suitable and they would probably give you the marks for that, if you backed it up with the realistic calculation
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    Yeah you see why I was confused now haha I also didn't notice that there was a second page on the thread for a while
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    Neither did I when I first replied, thought it hadn't gone through or something until I saw the second page! I'm just thinking though, why are we using 12V/10V as the voltage in the calculation when the voltage across the whole system is 24V due to it being +12V and -12V? Do you just always use the positive voltage?
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    The speaker is connected from +12V to 0V or from -12V to 0V depending on which FET is conducting so the maximum voltage it can ever have dropped across it is 12V. The + and - are there to cater for positive and negative input voltages. because the maximum voltage across the speaker is only 12V the maximum rms power output can only be rms of that 12 volts divided by the resistance of the speaker. Does that answer your question ?
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    Sorry, thought I'd replied to this! But yes, that makes sense, thanks! I think I had sort of assumed in my head that both could conduct at once, meaning 24V across the speaker but thank you for clarifying!
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    So how did everyone find the Unit 2 paper? I quite liked it really
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    It was alright, thought some of it was slightly odd
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    What was the answer to the wordy bit about the capacitors in series?
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    What were the answers to the two bandwidth graph questions ?. I also messed up one part of the RC combination section by getting my correct answer x 0.37
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    0.37 rings a bell actually...I can't remember exactly what I did but 2.2/6 was 0.37 so I used that at some point! As for the capacitors in series I'm not sure but I think it was probably something to do with the capacitance total being lower due to:

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    I think in total it was very peculiar, I was expecting boolean algebra, karnaugh maps, push-pull amplifiers and MOSFETs but all I got was weirdly worded questions and the easy voltage division!
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    Yeah, thought exactly the same as you - didn't expect the gain vs freq. graph's either...
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    For the gain question where we had to draw a curve how big was yours ? my one was pretty low. for the power one where we had to estimate I just wrote 10^3 to 10^4. couldn't work it out
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    It was a Gain Bandwidth Product graph so it was just a straight line going from 10^6 on the y-axis to 10^6 on the x-axis. As for the power one, the bandwidth is where the power gain is 50% of the maximum/required power gain, therefore you draw a horizontal line at 1.25W as the maximum power was 2.5W, then you read of the x values from where your straight line hits the curve, I can't remember what my answer was but that is how to do it
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    oh dear I don't think that test went very well at all for a start my max power was 3.5 W
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    For the calculation it was 3.5W so you've got that one right, the graph was for a different power output though unrelated to that calculation question.
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    ah cool cheers buddy
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    I think this explains the weirdness of our paper, in January a new electronics A level examiner person was advertised. I guess the paper was written by someone new! (See attached image)
 
 
 
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