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So I was solving a CIE Physics (9702) past paper and came across this question: [9709/41/O/N/09 // Q.7.B.]

9709/41/O/N/09 // Q.7.B.

The questions were:

7.(b).(i) On the axes of Fig. 7.2, draw a graph to show the variation with time t of the potential difference across diode A.

7.(b).(ii) On the axes of Fig. 7.3, draw a graph to show the variation with time t of the potential difference across diode B.

I did draw the graphs such that, for 7.(b).(i), when the alternating voltage is positive in figure 7.2, the voltage across diode A is also positive and when the alternating voltage is negative in figure 7.2, the voltage across diode A is 0, essentially drawing a half-wave rectification graph.

And for 7.(b).(ii), in figure 7.3, I drew the same graph but with a phase shift of 180 degrees showing that when the alternating voltage is positive the voltage across B is 0 and and when the alternating voltage is negative the voltage across diode B is positive.

Now the problem is that apparently this is wrong and the graphs will be drawn the other way around i.e. the graph that I've drawn for question 7.(b).(i) was actually the graph that was supposed to be drawn for 7.(b).(ii) & similarly the graph that I've drawn for question 7.(b).(ii) was actually the graph that was supposed to be drawn for 7.(b).(i). I can't seem to understand why it'll be this way though.

The examiner's report says this:

Question 7(b)

(i) Many candidates did not seem to realise that, for an ideal diode that is forward biased, the potential difference across it would be zero. There were many poorly-drawn sketches with peak values shown above those on Fig. 7.2.

(ii) There were very few correct responses here. In many answers, an inverted voltage was indicated.

I don't understand his explanation for it either.

The guy/girl from physics-ref.blogspot (He/She makes solved solutions for 9702 papers and more. Highly appreciate his/her help.) also answered this question along with an explanation for it. Their answer is correct but I didn't really understand what he/she meant by their explanation for it so if one of you readers do understand please try explaining it to me in simpler words I guess.

Here's his/her solution: enter image description here

Please help soon my exams are in less than a month. Thanks.

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    \$\begingroup\$ My immediate question when I came to check the answers was 'voltage with respect to what?' The graphs are given with respect to zero, but there is no ground connection shown on the question. This must mean the voltages have to be interpreted as differential. I assume the input voltage is defined as '+ve up', that's a common convention. I assume diode A anode is to be treated as +ve. Diode B is the other way up (geometrically), so should the graph be oriented for geometry, or anode? Who knows, surely no markdown for polarity. If you rigorously state your assumptions, you should be OK. \$\endgroup\$ – Neil_UK Sep 25 '16 at 6:37
  • \$\begingroup\$ An ideal diode, when it is conducting, behaves as an ideal wire. What is the voltage across an ideal wire? \$\endgroup\$ – user253751 Sep 25 '16 at 7:16
  • \$\begingroup\$ Alternatively: What is the voltage on each side of the diode? What do you get when you subtract those two voltages? \$\endgroup\$ – user253751 Sep 25 '16 at 7:16
  • \$\begingroup\$ @immibis I got the ideal wire concept though I don't understand what happens to the V in Diode A the very moment after the positive cycle changes and becomes negative. Negative V can't pass through Diode A so the V across it will still be 0 right? \$\endgroup\$ – Ishan S Sep 25 '16 at 7:29
  • \$\begingroup\$ @immibis I understood why the V is 0 now I need to understand where the bumps of positive potential come from, in the graph of V against T. \$\endgroup\$ – Ishan S Sep 25 '16 at 7:31
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Your answer was indeed backwards. When the AC voltage is positive, diode A is on. That means diode A's current is positive and its voltage is zero. When the AC voltage is negative, diode A is off -- the current is zero and the voltage is negative.

The schematic for this question is poorly-drawn because it does not show the polarities of the voltages.

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  • \$\begingroup\$ Thanks. the first half of your answer makes sense. However you're saying that When the AC voltage is negative, diode A is off -- the current is zero and the voltage is negative , if the voltage were negative wouldn't the graph show it? if you see the worked solution by the guy who solved in the image below in my post you'd see the voltage through either diodes is never negative. \$\endgroup\$ – Ishan S Sep 25 '16 at 7:21
  • \$\begingroup\$ Also how would I use what you said to draw a graph for the voltage through diode B. \$\endgroup\$ – Ishan S Sep 25 '16 at 7:21
  • \$\begingroup\$ +1, especially for highlighting the missing polarities (or "reference directions" of voltages as others call them). I'm fed off seeing lots of high level papers talking about electrical quantities quantitatively without specifiying directions. It's like saying we have 20°C without specifying if below or above zero. In this latter case common sense may help, but while explaining things to students common sense is, by definition, missing (you still have to develop it!). \$\endgroup\$ – Lorenzo Donati supports Monica Sep 25 '16 at 7:24
  • \$\begingroup\$ @IshanS The polarity of the diode voltage is not given, so either way is valid. When the diode is off, the forward voltage (from anode to cathode) is negative and the reverse voltage (from cathode to anode) is positive. I used the forward voltage. It looks like the Physics Ref person used the reverse voltage. \$\endgroup\$ – Adam Haun Sep 25 '16 at 17:17
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WHen measuring the output , you get the the half wave forward biased positive voltage and the diodes read 0.

When off the diode read negative when measured across the diode differentially with positive on anode of A.

Since this is a different reference point from your assumption, does that make sense now?

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  • \$\begingroup\$ Thank you for responding so fast. I'm sorry but it does not. I'm still confused why will the graphs be the other way around? \$\endgroup\$ – Ishan S Sep 25 '16 at 6:36
  • \$\begingroup\$ @IshanS Don't think of "why will the graphs be the other way around?". Think of "why is the voltage zero when the diode is conducting and negative when it is not?". \$\endgroup\$ – user253751 Sep 25 '16 at 7:17
  • \$\begingroup\$ if you probed the diode using two probes in differential mode, using the Anode as - and cathode as + now it will be inverted from the assumed reference in the question and will be positive half wave. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 25 '16 at 14:06

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