How do I calculate the current in a diode?

In summary, the question is to calculate the current in a diode with a forward voltage of 0.62V and a reverse saturation current of 10pa. The Ideal Diode Equation is used, where I = Is[e^(qv/kt) - 1]. The letter 'e' represents the base of the natural logarithm, which arises naturally in various contexts. Assuming room temperature, q/kT has a value of approximately 40V-1. For voltages greater than 0.1V, the 1 term in the equation can be ignored.
  • #1
vodkasoup
31
0

Homework Statement



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."



Homework 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.

The Attempt at a Solution



As above, I'm completely confused here.


Many thanks for your time and replies!
 
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  • #2
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.
 
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  • #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!
 
  • #4
I'm still not sure how to go about solving this. Am I on the right track with the equation I mentioned?
 
  • #5
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.
 
  • #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.
 
  • #7
vodkasoup said:
Well according to the notes I've been given, at normal room temperature q/kT has a value of approximately 40V-1.

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.

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

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]
 
  • #8
Got it! I actually didn't realize 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.
 
  • #9
Hey, sorry, I was traveling 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.
 
  • #10
Ahhh, it all makes sense now. Couldn't have done it without you sir, many thanks indeed!
 

Related to How do I calculate the current in a diode?

1. How does a diode affect current flow?

A diode is an electronic component that allows current to flow in only one direction. This means that when the diode is forward-biased, current can flow through it, but when it is reverse-biased, current is blocked.

2. What factors affect the current in a diode?

The current in a diode is affected by the voltage applied across it, the resistance of the diode, and the temperature of the diode.

3. How do I calculate the current in a diode?

The current in a diode can be calculated using Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R). In the case of a diode, the resistance is not constant and changes based on the voltage applied, so a more accurate calculation can be done using the diode's characteristic curve.

4. What is the diode characteristic curve?

The diode characteristic curve is a graphical representation of the relationship between the voltage applied across a diode and the current that flows through it. It typically has an exponential shape, with a small current flowing when the diode is reverse-biased and a rapidly increasing current when it is forward-biased.

5. Can I use a multimeter to measure the current in a diode?

Yes, a multimeter can be used to measure the current in a diode, but it must be done while the diode is in a circuit and under the correct biasing conditions. The multimeter should be set to measure current and connected in series with the diode to measure the current flowing through it.

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