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incremental resistance of a diode

Discussion in 'Electronics Homework Help' started by kbcheong, Mar 24, 2012.

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  1. kbcheong

    kbcheong

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    Mar 24, 2012
    A vacuum diode's voltage current characteristic is closely approximated by the Childs - Langmuir Law , with one parameter P called the perveance

    ip: Pv(PK)exponential3/2 if v(PK) > 0
    0 if v(PK) < 0

    A 6AL5 twin diode has P about 2.0mA /v exponential 3/2

    What is the incremental resistance (in Ohms) for the bias voltage of V(PK) = 4.0 V?

    I've calculated that the current Ip (in Amperes) is 0.016. I wish to get a formula to calculate the incremental resistance
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Resistance is R=U/I.
    Incremental resistance is accordingly dR=dU/dI

    Since usually you don't have an expression (formula) for U=f(I), you take the formula I=f(U) and calculate the derivative di/du=f'(u) which is 1/dR. From 1/dR it's easy to get R.

    Harald
     
  3. engrzubair

    engrzubair

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    Mar 27, 2012
    wt will be the current when Vpk is 10 volts kindly write current formula....and how to find incremental resistance plz if u know any formula then write it to me by the way R=U/I wt is U...???
     
  4. engrzubair

    engrzubair

    2
    0
    Mar 27, 2012
    i did like this....0.002*4*exp3/2=0.0064 not 0.016....exp3/2=4.4816
    so how u calculate th current
     
  5. Raven Luni

    Raven Luni

    798
    8
    Oct 15, 2011
    I see youre having trouble with that question as well.

    It makes no sense. I've tried every permutation of the numbers calculated for the slope and it doesnt work. Theres definitely something that hasnt been explained properly here.
     
  6. jackorocko

    jackorocko

    1,284
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    Apr 4, 2010
    Oh this is that MIT course?
     
  7. Raven Luni

    Raven Luni

    798
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    Oct 15, 2011
    It is. TBH I dont think they've explained any of the diode stuff particuarly well. I had trouble with an ealier question and just managed to get it. This one is bad. I've been through the whole lecture sequence again and at no point have they explained what incremental resistance actually is. All I know is that I have a voltage / current relationship, have the correct voltage and current for a certain point, but the checker says I'm wrong when I try to use those values to get resistance for that point.


    Actually - they havent explained diodes at all. In the last couple of sections theyve just thrown them in with the questions :p
     
    Last edited: Apr 9, 2012
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,490
    2,832
    Jan 21, 2010
    It sure is.

    Go back to the lecture and look at how the small signal model is created based on the original function for the curve and the Taylor expansion of it.

    This is a simple problem and if you don't get it you're in for a world of trouble...

    I find that the homework often precedes the lectures in some aspects. Doing the required reading may help. And it also may help to look at the beginning of the following lecture series.

    I can't remember how I did this question myself, there are at least 3 ways to do it. One is graphically (check where the lecture talks about treating a small part of a curve as a straight line), another way uses small deltaX and calculated deltaY values to calculate a value close to that graphical result. The third is the strict mathematical way of getting the slope of a line.

    In all cases there is something else to do because you may get a value that is conductance rather than resistance. (and I suspect this is your problem RavenLuni)

    If you simply search for the homework name as a keyword on the forums (at MITx) you'll find lots of help. (search for H4P0 and you'll get all the help you should need -- and maybe more)
     
  9. vikinggold

    vikinggold

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    Apr 9, 2012
    Diode incremental resistance

    Hi, I have been having a similar problem with this question on the MITx course. Maybe someone can offer some help with how to get the formula for this calculation. After searching for incremental resistance on the internet for some time all I could find were questions with given answers, but no explanation of how to get from one to the other. In the end I did arrive at the correct answer by examination of the sample answers that I found, but I would like to know how to get there properly.

    R(inc) = 2/3*(i(p)*V(pk))

    I have no idea where the 2/3 multiplication factor comes from, can anyone help. I know it gets the correct answer but I can;t find any equations to support this.

    BTW I seem to find the same with the MITx course, with questions ahead of the lectures, I have just started this week and have had a struggle to catch up with the homework assignments. Couldn't get the Lab assignment to work out for this week, none of the curves I got were anything like the graph shown. Anyway let's hope week 5 is better.
    Thanks for any help offered.
     
  10. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I don't understand the solution in its completeness, because i arrive at a different result. Only the factor 2/3 can be found in my solution:

    Start with:
    I=P*V^(3/2) (btw: that's the child-langmuir law, there is no exponential in the equation as wrongly stated in the first post).
    where I = anode current, V=anode voltage, P=P(V) and thus P depends on the operating point (defined by V).
    SInce a resistance is defined by R=V/I, I'd change the equation such that V=f(I):
    I/P=V^(3/2)
    (I/P)^(2/3)=V or V=(I/P)^(2/3)
    Now since we are interested in the incremental (or as i understand it differential) resistance, instead of R=V/I we need r=dV/dI (that is because we only want no know the increment of current caused by an increment of voltage (http://en.wikipedia.org/wiki/Electrical_resistance_and_conductance).
    This derivative is comparatively easy calculated from the last equation:
    dV/dI=(2/3)*(I/P)^(2/3-1)
    dV/dI= (2/3)*(I/P)^(-1/3)
    dV/dI=2/3*1/((I/P)^(^/3))
    This explains the factor 2/3 but not the rest of the equation.

    Harald
     
  11. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Start with: I=P*V^(1.5)

    The general form for the derivative is d/dx U^n = n*U^(n-1)U'

    d/dV I = d/dV P*V^(1.5) = P d/dV V^(1.5) = 1.5P*sqrt(V)

    Incremental Resistance = dV/dI = 1/(1.5P*sqrt(V)) = 2/3 * 1/P * 1/sqrt(V)
     
  12. Raven Luni

    Raven Luni

    798
    8
    Oct 15, 2011
    You know - the crazy thing is I'm sitting here staring at the main board for my MUNTER project (Z80 computer) and I'm thinking how the hell did I put all this together after this course has shown me how much I actually DONT know :p
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,490
    2,832
    Jan 21, 2010
    Remember all those levels of abstraction in the first lecture sequence?

    You don't need to understand everything that goes on in the abstraction levels below the ones where you're working. For example, how many people are experts at working their iPod, but really don't know an awful lot about digital electronics design.

    The same thing happens when designing something with digital electronics. You can get away with treating them as black boxes. However, knowing the underlying stuff will help you to make sense of, and apply, the things you read in the datasheet.

    Just for fun , I asked my wife's cousin (who has been a mechanical engineer for about 10 years now) "When was the last time you had to solve a differential equation"? His answer was "I think it was in second year at university".

    I tell you that as the course comes up to differential equations... :D
     
  14. kavi_11

    kavi_11

    1
    0
    Oct 2, 2012
    Answer

    Put the values of I and P in this formula..

    (dV/dI)= [2*(0.037exp(-1/3))]/[3*(0.002exp(2/3))]

    i also did like this and the answers were correct..
     
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