Calculating the voltage drop on a diode in series with a resistor

Thread Starter

subatomic particle

Joined May 8, 2018
65
Hello Guys,
so i was simulating the circuit below and i am wondering why is the voltage drop on the diode not the same as on the resistor, as i made the ohmic resistor of the diode equals to the resistance of the resistor?
Can you please explain to me how i can determine the voltage drop on a diode?
1656334677913.png
 

BobTPH

Joined Jun 5, 2013
5,248
That simulation cannot be correct. The voltage drop on the diode should be in the range of 0.6 to 0.7V.

You determine the voltage drop by looking at the IV curves in the data sheet. It is not a resistor, Ohm’s law does not apply.
 

Thread Starter

subatomic particle

Joined May 8, 2018
65
You cannot measure the ohmic value of a Diode using a VOM on the ohms range.
The diode require a forward bias current in order to measure the volt drop across it.
So in my circuit there is not way to calculate the voltage drop, since i dont have a bias current? how does the program decide whicch voltage is going to drop there?
 

PeteK

Joined Jul 31, 2019
1
Hello Guys,
so i was simulating the circuit below and i am wondering why is the voltage drop on the diode not the same as on the resistor, as i made the ohmic resistor of the diode equals to the resistance of the resistor?
Can you please explain to me how i can determine the voltage drop on a diode?
View attachment 270243
Hello,
there is a nice paper how to calculate the voltage drop on a diode in several ways, https://ece.uwaterloo.ca/~mhanis/ece241/lectures/set7.pdf
 

dl324

Joined Mar 30, 2015
14,700
Welcome to AAC! I see that you lurked several years before making your first post.
there is a nice paper how to calculate the voltage drop on a diode in several ways, https://ece.uwaterloo.ca/~mhanis/ece241/lectures/set7.pdf
Nice.

I wonder why teachers don't put dates and their names on their lecture materials? This from circa Q3 2004.

Moderators/administrators of this forum prefer to have all documents and images uploaded to the site so broken links don't become a problem in the future. File broken into 3 parts due to file size limit for uploads.
 

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Audioguru again

Joined Oct 21, 2019
4,798
A diode is not a resistor. The datasheet for a diode shows its forward voltage at different currents and at different temperatures.
A small silicon diode like a 1N4148 has a forward voltage of about 0.7V at 5mA at room temperature.
 

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SamR

Joined Mar 19, 2019
4,301
The voltage drop depends on the doping levels of the PN junction and its barrier potential and can vary from ~0.6-0.7 for silicon. There is no resistance value given in the PDF for a diode. There is a reactance value Zz given for Zeners. The Voltage to Current curve is so steep that the resistance is basically nil.
 
Last edited:

Jony130

Joined Feb 17, 2009
5,351
Hello Guys,
so i was simulating the circuit below and i am wondering why is the voltage drop on the diode not the same as on the resistor, as i made the ohmic resistor of the diode equals to the resistance of the resistor?
Can you please explain to me how i can determine the voltage drop on a diode?
View attachment 270243
Since the diode is a nonlinear element, we are unable to find the algebraic solution. We are forced to use numerical techniques. For example the iteration.

Since your diode has Is =1fA and n = 1 and assuming T_junction = 27°C we have Vt = 25.865mV.
We can try to guess the diode forward voltage.
The diode current relation is:

Id = Is*e^(Vd/Vt - 1)

Or the voltage drop across the diode is equal to:

Vd = Vt* ln(Id/Is +1) ≈ Vd = Vt* ln(Id/Is)

Because in the simulation you have set the diode ohmic resistance equal to Rd = 1kΩ. In fact, we have two 1kΩ resistors in series with the diode. Rd and R1.

So the first iteration I assumed Vd = 0.6V thus:

Id = (10V - 0.6V)/2kΩ = 4.7mA (1)

Now I plugging this value into Vd equation

Vd = 25.865mV * ln( 4.7mA/1fA) = 0.7547V (1)

And the second iteration

Id = (10V - 0.7547V)/2kΩ = 4.622mA (2)
Vd = 25.865mV * ln (4.622mA/1fA) = 0.7542V (2)


third iteration

Id = (10V - 0.7542V)/2kΩ = 4.622mA (3)

Since Id(2) ≈ Id(3) we end the iteration here.

Therefore, the final result is:

Vd = 0.7542V and Id = 4.622mA. But since you set Rd = 1kΩ in your circuit V_D = Vd + Id*Rd = 0.7542V + 4.622mA*1kΩ =5.3762V
 

k1ng 1337

Joined Sep 11, 2020
584
If you can, I recommend placing a silicon diode in series with different value resistors (like in post #14) and measure the voltage drop across each component. Plot the results on a graph and see if you can derive the relationship.
 
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