Calculating Diode Current with Is=50nA and Vd=0.6V | Homework Help

In summary, the question is asking for the diode current with an applied forward bias of 0.6V and Is=50nA. The equation used is Id = Is ( e ^ (kVd/Tk) - 1). After some confusion with the constants, the correct calculation is done to get an answer of 10.077nA. The next part of the question asks for the diode current at 20 degrees C for a silicon diode with Is=0.1uA at a reverse bias potential of -10v. The same equation is used with the given values, resulting in a diode current of -0.00335nA.
  • #1
HebrewHammer
8
0

Homework Statement



Determine the diode current with Is=50nA and an applied forward bias of 0.6V

Homework Equations



Id = Is ( e ^ (kVd/Tk) - 1)

The Attempt at a Solution



So my teacher is horrible and this is what I have been able to collect on my own.

Is = 50 and Vd = 0.6v
I'm pretty sure k is referring to Boltzmann’s constant k = 1.38 x 10^-23 J.K-1.
I'm pretty sure Tk is the room temperature in kelvin so that would be 298.

Plugging it all in I get
Id = 50 ( e ^ (((1.38 x 10^-23)(.6))/298)-1)

This gives me an answer of 1.389*10^-24
Something seems really off, if one of those constants I figured out is wrong please let me know.
 
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  • #2
The exponential term should actually be

e ^ [e·Vd / (k·T)]​

where e is the charge of an electron, k is Boltzmann's constant, and T is absolute temperature. (Vd is the diode voltage, of course.)

Also, you seem to be using 50 Amps for Is, when it is actually 50 nA.
 
  • #3
Ok so I figured out how to do it.

50nA(e^(.6/298)-1) =

50nA(.002015) =

10.077nA

Seems like I was overthinking it. Now there's a part 2 to this question that I am really lost on.


Determine the diode current at 20 degrees C for a silicon diode with Is=0.1uA at a reverse bias potential of -10v.

Further help would be awesome.
 
  • #4
HebrewHammer said:
Ok so I figured out how to do it.

50nA(e^(.6/298)-1) =

50nA(.002015) =

10.077nA

Seems like I was overthinking it.
There's still a problem. You have completely left e and k out of your calculation!

Now there's a part 2 to this question that I am really lost on.

Determine the diode current at 20 degrees C for a silicon diode with Is=0.1uA at a reverse bias potential of -10v.

Further help would be awesome.
You are given Vd and the temperature, so just apply the same equation. (Don't forget about e and k.)
 
  • #5
ok i got the answer i think

.1(e^(-10/293)-1) =

= -.00335nA

Seems off to me but its the same procedure.
 
  • #6
There's still a problem. You have completely left e and k out of your calculation!

The exponential term should actually be
e^[e·Vd / (k·T)]​
where e is the charge of an electron, and k is Boltzmann's constant.
 

What is a diode?

A diode is an electronic device that allows current to flow in only one direction. It is made up of a semiconductor material, typically silicon, with two terminals – an anode and a cathode.

How does a diode work?

A diode works by utilizing the properties of the semiconductor material it is made of. When a voltage is applied in the forward direction (from anode to cathode), the diode allows current to flow. However, when a voltage is applied in the reverse direction, the diode blocks the flow of current.

What is diode current?

Diode current refers to the amount of electric current flowing through a diode. It is measured in units of amperes (A).

How do you determine diode current?

The diode current can be determined by applying a voltage across the diode and measuring the resulting current flow. This can be done using a multimeter or other measuring device. It is also important to consider the voltage and current ratings of the diode to ensure it is not damaged by the applied voltage.

What factors affect diode current?

The main factors that affect diode current include the voltage applied, the resistance of the circuit, and the characteristics of the diode itself (such as its forward voltage drop and reverse leakage current). Temperature can also affect diode current, as higher temperatures can increase the leakage current and decrease the forward voltage drop.

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