1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Diode circuit problem

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A moving coil ammeter is to be adapted to detect small alternating currents. Which of the diagrams shows how a diode could be connected in order to make the conversion ?

    (Options shown in attachment)

    (Ans: (3))

    2. Relevant equations



    3. The attempt at a solution
    I am not sure if I understand the problem. What does it mean by "detect small currents"? Also, it doesn't state if the diode are considered to be ideal or not? But I think I am supposed to consider them as ideal.

    There are my thoughts:

    Circuit (1): The diode is non-conducting if + at the upper terminal and - at the lower terminal, so all the current pass through the ammeter. If the signs are switched, no current passes through the ammeter as the diode is now conducting.

    Circuit (2): Opposite to the functioning of circuit (1).

    Circuit (3): If + at the upper terminal and - at the lower terminal, the diode is conducting and ammeter records a reading. If the signs are switched, the circuit gets open and no current through ammeter.

    Circuit (4): Opposite to the functioning of circuit (3).

    But I am not sure which option to select as answer. :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 2, 2014 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    If you picked (1) or (2) and the diode requires at least 0.6V to conduct appreciable current, where would a current of amplitude I_0 go if the ammeter resistance is R?
     
  4. Apr 2, 2014 #3
    To the ammeter? :uhh:
     
  5. Apr 2, 2014 #4

    Curious3141

    User Avatar
    Homework Helper

    Straight away, you can throw out the parallel arrangements in (1) and (2).

    The main purpose of the diode is to rectify the AC into DC.

    If the supply voltage is lower than 0.6V in amplitude at any point in time, then even in forward-bias, all your current is going to go across the ammeter. In reverse-bias, the same holds true, of course. In other words, the diodes are doing nothing in (1) and (2).

    Even if the supply voltage is always higher than 0.6V in amplitude (and therefore the diode in parallel is forward biased 50% of the time), the diode will not short out the ammeter. Remember that ammeters have low resistance (ideally zero), so you will simply have a situation where part of the current goes through the diode and the rest through the ammeter. That's 50% of the time. The remaining 50% of the time (when the diode is in reverse bias), all the current goes through the ammeter.

    Either way, you've not achieved the aim of rectification. You still have an AC signal passing through the ammeter instead of a varying DC signal. You will just have "flutter" of the needle.

    It's easy to rule out (4) because the ammeter will always read zero, whether the diode is in forward or reverse bias.

    (3) is the only correct option. 50% of the time, the diode is forward-biased, and the ammeter polarity is correctly oriented to measure the current. The remaining 50% of the time, the diode is in reverse bias, and no current passes through the ammeter.

    What usually happens in a DC ammeter is an "averaging" of a varying DC currrent. So you're measuring the average value of a half-rectified waveform. As an exercise, can you derive what current you're measuring (with respect to the RMS current) for a perfectly sinusoidal input?

    The better design is a bridge-rectifier, which gives you full rectification. This way, you will measure a value much closer to the true RMS current.
     
  6. Apr 2, 2014 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Right. It would go in BOTH directions, not just one, so the poor little ammeter would try to swing its little needle back and forth at 60 Hz (which it can't do so it would look like a blur on the meter).

    So 1 and 2 would seem a poor choice, would it not?
     
  7. Apr 3, 2014 #6
    Hi rude man and Curious! Sorry for the delay in reply.

    Yes but the question doesn't state anything about rectification. It simply asks about the situation where ammeter "detect small current". I am still not able to understand what is the meaning of "detect small current". :(

    Just to confirm if I am understanding your above explanation, please have a look at the attachment. :)

    (R is the resistance of ammeter.)
    Again, why rectification? :confused:

    Do I have to calculate this for circuit (3)?
     

    Attached Files:

  8. Apr 3, 2014 #7

    Curious3141

    User Avatar
    Homework Helper

    I'm assuming the DC ammeter is one that can only read currents going in one direction (from the '+' to the '-' terminal). There are DC ammeters (zero-centred) that can read currents going both ways, but this complicates the analysis a bit. (3) is still the best layout, but (4) becomes quite acceptable as well (it's just that the needle will always swing only to the negative side of the scale).

    I'm making another assumption. The frequency of the input AC current is high enough that if you do nothing to rectify it, you won't really see a steady swing from peak current to zero and back. Either you will see high frequency flutter about a mean position, or the needle will just appear steady at the mean current. This is actually a fair assumption, implicit in most of these questions. Moving-coil ammeters usually are incapable of reading AC currents, even if they're zero-centred.

    In order to measure the current from an AC source, you need to do something so that it flows only one way (and the right way, from '+' to '-') through the ammeter. The way to do that is rectification. You can look up half-wave and full-wave rectification (and smoothing) to get a better idea of what's involved. But with just one diode, half-wave rectification is all you can achieve.

    You don't actually have to calculate the average current (btw, scratch the "RMS" mention in my post - I think the average is what's measured). That was just meant as a thought-provoking exercise.

    As far as your graph goes, let's make some simplifying assumptions (I did!). The ammeter has zero resistance throughout. The diode has zero resistance in forward bias above a voltage of 0.6V and infinite resistance in reverse bias (or when forward biased with a voltage of less than 0.6V). The input voltage is a perfect sinusoid obeying ##V(t) = 2\sin t##. The source load is exactly ##1 \Omega##, so the current flowing from the source is exactly that, too, without a phase shift.

    Given that, how do you think the graph should look in the parallel arrangements (1 or 2)?

    I think it should go something like this (see attachment). The blue curve represents the input current (and voltage), while the red one represents the current measured by the ammeter (though the ammeter will tend to "average" out the varying current, as I mentioned). I think of it as a "Batman" shaped curve for the first half of the cycle, and a flat line for the second. Can you explain why?
     

    Attached Files:

  9. Apr 3, 2014 #8
    Thank you for the detailed response!

    My thoughts are that when ##t=\arcsin(0.3)##, the diode starts conducting and 50% current goes to each of the branches till the diode is conducting. When the sign of input waveform changes, the DC ammeter won't conduct any current for the reasons mentioned in the first sentence of your post.

    Did I interpret correctly?
     
  10. Apr 3, 2014 #9

    Curious3141

    User Avatar
    Homework Helper

    Exactly.:approve: Very idealised and unrealistic, but good to work through it like this, I think.

    But you shouldn't need to do all this to see that (3) is the most appropriate answer.
     
  11. Apr 4, 2014 #10
    Won't the output for (3) be similar to (2)? :confused:

    For circuit (3), at ##t=\arcsin(0.3)##, the diode starts conducting and till time ##t=\pi-\arcsin(0.3)##, the ammeter reads a reading of ##2\sin(t)-0.6##. For the second half cycle, the ammeter reads zero. For ##0\leq t \leq \arcsin(0.3)## and ##\pi-\arcsin(0.3)\leq t\leq \pi##, the ammeter in (2) reads some reading whereas in (3), the ammeter reads zero but how does it helps to select the correct answer? :confused:
     
  12. Apr 4, 2014 #11

    Curious3141

    User Avatar
    Homework Helper

    The ammeter measures *current* not voltage. Aren't the diode and ammeter in series? Isn't the same current passing through them (with the diode in forward bias)? :wink:
     
  13. Apr 4, 2014 #12
    When did I say it measures voltage? :confused:

    I was using your setup where the resistance is ##1\,\Omega##.

    Yes, the same current passes through both but how does that answer my question? :confused:
     
  14. Apr 4, 2014 #13

    Curious3141

    User Avatar
    Homework Helper

    Sorry, you reckoned correctly based on my setup, but this is still better than the parallel arrangement which is reading at most half the current, isn't it?

    What if the input wasn't 2sin(t), but something like 10sin(t)? The 0.6V dropped across the diode would not affect the measurement as much. With the series arrangement, the current measured at the peak would be much closer to the true value, whereas with the parallel arrangement, it would still be half.

    Sometimes with idealised problems, you may assume there is no voltage drop across a diode even in forward bias - i.e. as long as it is in forward bias, it is always conducting with zero resistance (it still doesn't conduct in reverse bias, i.e. reverse bias implies infinite resistance). In which case, the arrangement in (3) will give you a perfect reproduction of the input current, whereas (2) would have the same issue.

    I should have idealised it to that level to make things clearer.
     
    Last edited: Apr 4, 2014
  15. Apr 4, 2014 #14

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Just to butt in again for a moment - this may be redundant since I haven't read all the posts completely - but I get the impression we're applying voltages to this thing. You can't apply voltages directly to a galvanometer - at least not larger voltages.

    Config. (3) will measure current accurately, but it's the current with the ammeter in situ. If you have a voltage V and a series resistor R it will not measure V/R. It will measure V/R quite closely only if both V and R are sufficiently large. The diode represents a non-linear (both a static and a dynamic) impedance in series with R, and the voltage has to exceed the diode drop by a healthy amount.
     
  16. Apr 5, 2014 #15
    Thanks a lot both of you! :smile:
     
  17. Apr 5, 2014 #16

    rude man

    User Avatar
    Homework Helper
    Gold Member

    OK Pranav - I added in bold a word to my last post.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Diode circuit problem
  1. Circuits with diode (Replies: 9)

  2. Circuit with diode (Replies: 4)

  3. Diodes in a circuit (Replies: 7)

Loading...