Oxford PAT 2014, Question 16

2014_page15

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6 thoughts on “Oxford PAT 2014, Question 16

  1. Hi,

    Could you explain the terms forward bias and reverse bias in relation to diodes and what the voltage drop means? I understand that diodes have a very high resistance in the ‘opposite’ direction, so what does breakdown voltage mean / relate to the shape of your graph, and why is the value at which the current approaches infinity, 0.6 V?

    Thanks!

  2. thank you very much for your reply.
    Can you please explain this point you’ve made? “Then you need to think about the abnormal case — a big discharge. This is another keyword: charge needs to flow to ground, so a big current needs to flow. From the graph we can see that the protection circuit allows any amount of current to flow but the voltage never exceeds the forward voltage drop, so the voltage across the amplifier inputs is limited and this protects the amplifier.”
    Thanks again!

    1. Sorry that wasn’t very clear, and it was maybe stretching a point. I meant that the word ‘discharge’ is a keyword, but that is maybe putting it a bit strongly. In any case a discharge (like a lightning discharge, for example) involves a sudden big current flow from a high potential to a low potential.

      If the discharge current goes through the amplifier it could destroy it. But in this case the diode protection circuit works like a lightning conductor would, providing a path of near zero resistance as long as the voltage across it is equal to the forward voltage drop of the diode: in this case a huge current can flow through the protection circuit, but the voltage across the amplifier inputs never exceeds the forward voltage drop of the diodes.

  3. Hello,
    What was your thought process when you try to tackle this kind of questions (when knowledge is applied in unfamiliar real life situation)? I have come across diodes before in AS, and I know their characteristics. However, when I need to apply my knowledge to these type of questions, I usually get really stuck. Do you have any technique that you use, and what should I do to improve on these type of questions?
    Thank you very much!

    1. Hi Claire, thanks for a very interesting question.

      In this case you are asked to sketch a graph and so that is the place to focus your energies, especially since the second part of the question sounds so weird, the reasoning being that the graph might shed some light on the second part. When sketching a graph of a physical situation there are a few techniques to remember.

      1. Try to work out the function
        If you are asked for a sketch see if you can work out the function that relates the quantities to be sketched. If you can, then sketch the function and you are done. This is obviously the best way to proceed, but in this case it’s hard to come up with a function to model the behaviour so this doesn’t work as a technique.
      2. Try some sensible points
        If you can’t think of a function then pick some values in the domain/range of the function and try to work out the corresponding values in the range/domain. Don’t forget zero, -∞, +∞ etc. Using this technique definitely buys us one point on the curve because it’s pretty obvious that when the voltage is 0 the current is 0. Better than nothing I suppose, but I doubt we’ve got a mark yet, and this graph is a bit nasty because the current is not defined for most of the possible voltages.
      3. Think about the slope of the graph
        If you are still stuck after thinking about value pairs, then try to think about the slope of the graph at any point, especially around any of the value pairs that you have worked out. In this case we know the curve goes through (0,0) so we can think about what happens when the voltage changes by some small quantity δV around the value 0.

        1. At this point you need to have a bit of knowledge and to know that diodes have a small forward voltage drop (this is the point at which a lot of candidates will get stuck, so if you know this then you are doing well). If you remember this you can see that the current through the forward biased diode will still be zero (and the current through the reverse biased diode will obviously be zero), so the net current is still zero. Using symmetry we can see we get the same thing for -δV. So we can start drawing a little horizontal line around the origin.
        2. The thing that everybody will most likely know is that (forgetting the forward voltage drop) a diode’s forward resistance is very low, so there needs to be a near-vertical line somewhere. If you remember the forward voltage drop, then you should be able to see that the line goes horizontally up to that voltage and then vertically after.

        So in this case, thinking about the slope, plus a bit of knowledge about diodes, gets us a long way.

      If we now think about the second part of the question there are some more techniques to consider.

      Look for keywords, especially adjectives
      Don’t forget the adjectives — they are always there to communicate some information. The amplifier is a ‘sensitive’ one for measuring ‘small’ signals. So we are considering small changes in voltage around the origin, and in this case we can assume that the signals are less than the forward voltage drop. In this normal case the protection circuit just looks like an extremely big resistor. Then you need to think about the abnormal case — a big discharge. This is another keyword: charge needs to flow to ground, so a big current needs to flow. From the graph we can see that the protection circuit allows any amount of current to flow but the voltage never exceeds the forward voltage drop, so the voltage across the amplifier inputs is limited and this protects the amplifier.

      And when you have solved a problem, look back over it and try to think about techniques and arguments that you used, and try to refine them so that you can apply them in similar circumstances.

      I hope this is some help. Finally, I know this is easy to say and harder to do, but think positively, stay calm, and concentrate on what you do know, not what you don’t.

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