Solve Simple Diode Problem: Voltage Drop @ 300K

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In summary, at a current of 73mA and a saturation current of 0.010mA, the voltage drop across the diode is 0.230V.
  • #1
Qbit42
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Homework Statement


You have a series circuit consisting of a dc power supply, a 90 ohm resistor and a diode.
If the measured current in the circuit is 136 mA, and the saturation current for the diode is 0.006 mA, what is the voltage drop across the diode? Assume the diode temperature is 300 K. Your answer should be given to three places after the decimal point.



Homework Equations


[tex]I = I_{o}(e^{\frac{qV}{k_{b}T}}-1)[/tex]

[tex] q = 1.602x10^{-19}[/tex]
[tex]k_{b} = 1.38x10^{-23}[/tex]


The Attempt at a Solution



Rearranging the above the equation I get

[tex]\frac{k_{b}T}{q}ln(\frac{I}{I_{o}} + 1) = V[/tex]

Plugging in the numbers I get 0.259V which isn't the right answer, I know this problem isn't difficult but I can't seem to wrap my head around why this doesn't work.
 
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  • #2
Qbit42 said:

Homework Statement


You have a series circuit consisting of a dc power supply, a 90 ohm resistor and a diode.
If the measured current in the circuit is 136 mA, and the saturation current for the diode is 0.006 mA, what is the voltage drop across the diode? Assume the diode temperature is 300 K. Your answer should be given to three places after the decimal point.



Homework Equations


[tex]I = I_{o} e^{\frac{qV}{k_{b}T}}-1[/tex]

[tex] q = 1.602x10^{-19}[/tex]
[tex]k_{b} = 1.38x10^{-23}[/tex]


The Attempt at a Solution



Rearranging the above the equation I get

[tex]\frac{k_{b}T}{q}ln(\frac{I}{I_{o}} + 1) = V[/tex]

Plugging in the numbers I get 0.259V which isn't the right answer, I know this problem isn't difficult but I can't seem to wrap my head around why this doesn't work.

I believe you are missing a pair of parenthesis in your first equation...
 
  • #3
berkeman said:
I believe you are missing a pair of parenthesis in your first equation...

Oh yeah I missed that, fixed it in the post above
 
  • #4
Qbit42 said:
Oh yeah I missed that, fixed it in the post above

So does that fix your answer?
 
  • #5
no I formulated my equation from the correct expression, I just mistyped it in the original post
 
  • #6
Your result is correct.

ehild
 
  • #7
Well that sucks, the problem is in an online assignment and it only accepts the right answer (we get 3 tries at it). It won't accept that one.
 
  • #8
Interesting that it gives the resistance of a series resistor and does not use its value in the question. Maybe, they wanted to ask the emf of the source.

ehild
 
  • #9
ehild said:
Interesting that it gives the resistance of a series resistor and does not use its value in the question. Maybe, they wanted to ask the emf of the source.

ehild

Well there is a second part to this problem where the resistance of the resistor is needed, I just didn't post it since it had no bearing on this part of the problem. Here it is:

What is the voltage across the diode-resistor combination?
 
  • #10
I have a few ideas about what might be wrong.

1. 0.006 mA seems pretty large for a diode saturation current. I think nA or pA range is more reasonable. Do they really say 0.006 mA, or could it have been 0.006 μA? Or something else?

2. Perhaps there are too many significant figures in the answer "0.259 V". Try the calculation using saturation currents of 0.0055 and 0.0065 mA, to get a sense of how many sig figs are justified.

3. Shouldn't the diode equation include an extra material-dependent ideality factor in the qV/kT term? Perhaps they expect the student to know what it should be, so they didn't provide the value in the problem statement.
 
  • #11
1. 0.006 mA seems pretty large for a diode saturation current. I think nA or pA range is more reasonable. Do they really say 0.006 mA, or could it have been 0.006 μA? Or something else?

I just copied and pasted the problem directly from the problem set

2. Perhaps there are too many significant figures in the answer "0.259 V". Try the calculation using saturation currents of 0.0055 and 0.0065 mA, to get a sense of how many sig figs are justified.

It says it wants my answer within 3 numbers after the decimal point

3. Shouldn't the diode equation include an extra material-dependent ideality factor in the qV/kT term? Perhaps they expect the student to know what it should be, so they didn't provide the value in the problem statement.

I took that equation directly from our lab handout, this question is a for a prelab problem worth a small fraction of the lab mark, but I also found it mentioned in the textbook (however nothing really much was said about it)
 
  • #12
Okay, sounds like those were three dead ends. Oh well, sorry.
 
  • #13
thanks anyways, I guess I'll just have to forget about it.
 
  • #14
You are doing it right, I had the same question but different values:
120 ohm resistor, current in the circuit (I) = 73mA, saturation current (Io) = 0.010mA, and T=300K.
My answer came out to be 0.230V which was accepted as correct. Just try it again and make sure all your calculations are correct.
 
  • #15
I'm not sure how to get the second part:
What is the voltage across the diode-resistor combination?
Any ideas?
 
  • #16
dfs730 said:
I'm not sure how to get the second part:
What is the voltage across the diode-resistor combination?
Any ideas?

Have you considered VD + VR ?
 
  • #17
Ahh...yes. I had put in Vr-Vd.
Vd + Vr worked, I got 8.99V/
Thanks!
 

FAQ: Solve Simple Diode Problem: Voltage Drop @ 300K

What is a diode?

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

What is the voltage drop across a diode?

The voltage drop across a diode is the difference in potential between its anode and cathode terminals. This value can vary depending on the type of diode and the current flowing through it.

How do you calculate the voltage drop at 300K?

The voltage drop at 300K can be calculated using the Shockley diode equation, which takes into account the diode's current, temperature, and ideality factor. It is given as V = VTln(I/Is), where VT is the thermal voltage, I is the current, and Is is the reverse saturation current.

What is the significance of 300K in calculating the voltage drop?

300K is the standard temperature at which most diode characteristics and parameters are measured and specified. It is equivalent to 27 degrees Celsius and is used as a reference point for diode operation.

What factors can affect the voltage drop across a diode?

The voltage drop across a diode can be affected by the diode's temperature, current, and the material it is made of. It can also be influenced by external factors such as load resistance and voltage supply. Additionally, the diode's age and manufacturing process can also impact its voltage drop.

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