Help finding voltage across diode

[Heylow]

Me and my friends have been working on this for days, and none of us are even close to the answer.

We are supposed to calculate the voltage drop (Vd) across the diode, while the saturation current is equal to 3x10^-16 A.
R1 = 1kΩ and
Is = 1 mA

The circuit can be found at:

Any help would be appreciated.

 

[Laplace]

There are some conditions left unsaid by the problem so I would just assume ideal conditions. Use the ideal diode law at room temperature, 300 Kelvin ( http://pvcdrom.pveducation.org/SEMICON/EQUAN.HTM ). Then for any given diode voltage you can calculate the corresponding diode current. To find the voltage across the resistor multiply the resistor value by the difference of the current source and diode current. For the next step I would recommend using a spreadsheet with three columns: Diode Voltage, Diode Current, Resistor Voltage. In the first column, increment the voltage over a range. Parameterize it so you can go first over a 1 volt range by 10 millivolt increments, then over a range of 0.1 volt (e.g. 0.5 thru 0.6 volt) by 1 mv increments, etc. In the second column, calculate the corresponding diode current. Then use the diode current to calculate the resistor voltage in the third column. Continue drilling down the increments. Where the voltage value in the first column equals the voltage value in the third column, that is your answer!

 

[barathbushan]

What is this saturation current ??
Does it flow through R1 or through the diode (reverse bias current ??) ,
but definitely it does not flow at the place you marked in the circuit diagram


CLUE TO SOLVING THE PROBLEM - - a current source in parallel with a resistor and a voltage source in series with a resistor , are interchangeable
, by using basic ohm's law equations . so convert the current source into a
voltage source [more easy to analyse], and solve the problem .

 

[Laplace]

The saturation current is a parameter of the diode itself, and is the "Io" value in the ideal diode law.

 

[ArFa]

Me and my friends have been working on this for days, and none of us are even close to the answer.

We are supposed to calculate the voltage drop (Vd) across the diode, while the saturation current is equal to 3x10^-16 A.
R1 = 1kΩ and
Is = 1 mA

The circuit can be found at:

Any help would be appreciated.

That is easy...
Just do the equivalent of current source and parallel resistor to Voltage source and series resistor .

Vin=i*R
R=R;

and then, when Vin>0 diodes is conducting and Vout is zero ( if diode is ideal), but if Vi<0 thn diode is cut off and Vout is Vin => Vout=Vin.

 

[lexroxas]

This is a very tricky question for a junior member to ask so please forgive me for providing the solution. I hope the moderator will understand.

1. Remove the resistor only. This makes the circuit a simple series circuit.
2. The equivalent thevenin voltage can now be computed using the formula Id = Is(e^(kV/Tk)-1). Id is the diode current and that is 1mA, k=11,600/2 for silicon, Tk =273+25, and V is the diode bias voltage and this is the thevenin equivalent voltage which is equal to 1.4815221 Volts.
3. The equivalent thevenin resistance is equal to Vd/Id and this is equal to 1.4815221 divided by 1 mA = 1,481.5221 ohms. This is the DC equivalent resistance of the diode and this is also the thevenin equivalent resistance.
4. Draw the thevenin's equivalent circuit and reconnect the 1k resistor in the thevenin equivalent circuit.
5. Solve the voltage drop across the 1k resistor. The drop is 1.4815221 Volts x(1k/(1k+1,481.5221 ohms) = 0.5970215216 Volt.

0.5970215216 Volt is the voltage drop across the 1k resistor at room temperature and this is also the voltage drop across the diode because they are in parallel in the original circuit. This is the drop at 25 degress only - room temperature and assuming that the diode is a silicon diode, if it's germanium then divide 11,600 by 1.

The actual diode current can now be recomputed from the saturation current formula above by using the voltage across the diode as 0.5970215216 Volt.

 

[(*steve*)]

This is a very tricky question for a junior member to ask

Remember that you're a junior member too.

Junior just means "not many posts", not young, silly, or in primary school. If this person was asked this question, one can assume that it was one which is appropriate given the contents of the course that they are doing.

 

[lexroxas]

Yes I definitely agree that I'm a junior. It's just when I read the transcriptions of this post, Laplace suggested the incremental method to solve the problem while Arfa suggested thevenizing the current source and the resistor. The inquirer was still probably lost with all the suggestions.