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!

Homework Help: How do I calculate the current in a diode?

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data

    The question I have been set is "Calculate the current in a diode when the forward voltage is 0.62V and the reverse saturation current is 10pa."



    2. Relevant equations

    I really have no idea how to begin solving this one. The only equation that I have been taught (aside from Ohm's Law) is 'the Ideal Diode Equation' - "I= Is [e qv/kt - 1]". However, I do not know what the 'e' in this equation represents and so do not know how to feed the numbers into the equation.

    3. The attempt at a solution

    As above, I'm completely confused here.


    Many thanks for your time and replies!
     
  2. jcsd
  3. Sep 26, 2010 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to PF,

    The letter 'e' is an irrational number. This number is the base of the natural logarithm. In other words, loge(e) = 1. Sometimes the natural logarithm is also written as ln, in which case we would say ln(e) = 1.

    Why is this number being chosen to be raised to some power, instead of some other number like 10 or 2? Because the exponential function ex arises naturally in a variety of different contexts and it has some unique (and very useful) properties. For instance, it is equal to its own derivative: d(ex)/dx = ex.

    Note, the exponential function is sometimes also written as exp(x), but this means ex.

    All of this is covered in any introductory calculus course. If you're doing a course in which you're trying to understand and model the properties of various electronic devices like diodes, then I'm very surprised that you wouldn't have come across this by now.

    Edit: if you click on the hyperlinked word "exponential" in my post, it takes you to the PF Library where there is more information on this topic. Neat.
     
    Last edited: Sep 26, 2010
  4. Sep 26, 2010 #3
    Thankyou cepheid for the welcome and your helpful reply :)

    I agree, it does seem strange that I don't know basics such as these - I've only just started this course after taking a break from education for several years, so it's probably a case of my forgetting lessons I was given in the distant past.

    I guess googling myself a crash course in solving exponential equations is the next step for me. :)

    Thanks again!
     
  5. Sep 26, 2010 #4
    I'm still not sure how to go about solving this. Am I on the right track with the equation I mentioned?
     
  6. Sep 26, 2010 #5

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah, you're on the right track. You've been given two of the parameters in that equation. The voltage across the diode is V. You can see from the equation that when V becomes large and negative, I = -Is. Hence, this is the reverse saturation current. I am guessing you are meant to assume something for the temperature.
     
  7. Sep 26, 2010 #6
    Well according to the notes I've been given, at normal room temperature q/kT has a value of approximately 40V-1.

    I assume that this all means that I = Is[e40v] , where Is is the reverse saturation current (10pa)?


    Thanks so much for your help here by the way.
     
  8. Sep 26, 2010 #7

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right, okay. There is no magic there. The quantities 'q' and 'k' are fundamental physical constants. The elementary charge, q, is the charge of an electron or a proton. The other constant, k, is the Boltzmann constant. So if you assume room temp and just plug in the values for the other two constants, that's where the 40 V-1 comes from.

    That's pretty close, but what happened to the -1 term in the equation?


    Thanks so much for your help here by the way.[/QUOTE]
     
  9. Sep 26, 2010 #8
    Got it! I actually didn't realise that to work out [e40v], all I had to do was press the 'ex' button on my calculator :redface: Like I said, it's been a while.....

    I have no idea where that -1 went; it seems to have become a casualty somewhere along the way. The notes given by my lecturer are rather cryptic at best -- he mentions that if V is greater than 0.1V , then I is given by I = Is[e40V]. The answer comes out correctly, although I don't necessarily understand how we arrived at it.
     
  10. Sep 27, 2010 #9

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hey, sorry, I was travelling to a conference. Oh yeah, it's simple. e4 is a pretty big number, much bigger than 1. So for high enough voltage, you can ignore the 1.
     
  11. Sep 27, 2010 #10
    Ahhh, it all makes sense now. Couldn't have done it without you sir, many thanks indeed!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook