I have used a diode having a knee voltage of around 0.72V. But in simulation, in a graph, I m not getting that value. is that because of the resistor

boostbuck

Joined Oct 5, 2017
109
Yes - the current in your sim is very small. The datasheet shows a 0.7 knee at If = 10mA, which in your circuit would be provided by a 2k resistor.
 

Thread Starter

valorous

Joined Oct 28, 2021
17
Yes - the current in your sim is very small. The datasheet shows a 0.7 knee at If = 10mA, which in your circuit would be provided by a 2k resistor.
thanx bro. Is datasheet given in Ltspice??
and also when i tried 2k resistor i m not getting correct result, so i used resistor of around 0.1k , and so i m getting correct value now.
 

Thread Starter

valorous

Joined Oct 28, 2021
17
thanx bro. Is datasheet given in Ltspice??
and also when i tried 2k resistor i m not getting correct result, so i used resistor of around 0.1k , and so i m getting correct value now.
and also is it necessary to show the knee voltage = 0.72V.
i mean, in my case, i m getting a diff value(in graph , where R=100k) for forward characteristics of diode , is that will be okay or not??
 

Papabravo

Joined Feb 24, 2006
17,646
You should not expect a simulation to match a datasheet exactly. The datasheet numbers should be treated as if they were normally distributed random variables with a mean and a variance. A datasheet value listed as a MINIMUM or a MAXIMUM should be interpreted as being 3 standard deviations away (+ or -) from the mean. Any particular part that you buy from a supplier has a 99.5% chance of being within ±3σ of the mean. A simulation model should also have the same 99.5% probability of being within ±3σ of the mean also. If you get really clever and decide to tweak the model(s) for your purposes like Alex ( @Bordodynov ) does from time to time you can adjust your result to your preference.
 
Last edited:

MrChips

Joined Oct 2, 2009
25,065
I don’t know if there is an exact definition of knee voltage.
Plot the current on a logarithmic scale and let’s see what you get.
You need to take the diode current beyond 10mA.
 

Papabravo

Joined Feb 24, 2006
17,646
is this count as a right or wrong, ??
and so, it is not necessary to go along with datasheet, right?
It counts as incomplete. You need to explore a much wider range of values. In particular you must extend the potential current to 10 mA and beyond. Like this:
At about 0.72V forward bias the current is in excess of 16.36 mA. At ≈12 mW of power dissipation it won't even get warm.

1635477284275.png

1635477343661.png
 
Last edited:

Thread Starter

valorous

Joined Oct 28, 2021
17
so when i am using resistor in ckt, i have to set the values such that it
It counts as incomplete. You need to explore a much wider range of values. In particular you must extend the potential current to 10 mA and beyond. Like this:
At about 0.72V forward bias the current is in excess of 16.36 mA. At ≈12 mW of power dissipation it won't even get warm.

View attachment 251408

View attachment 251409
so when i am using resistor in ckt, i have to adjust resistor value such that it explores wide range of values of current or to get the desired knee voltage(0.7V) , right?
 

Papabravo

Joined Feb 24, 2006
17,646
so when i am using resistor in ckt, i have to set the values such that it

so when i am using resistor in ckt, i have to adjust resistor value such that it explores wide range of values of current or to get the desired knee voltage(0.7V) , right?
That is not usually what you do. Normally you decide on what current you want in the diode and let the forward voltage be whatever it needs to be. In a career spanning half a century I've never seen somebody try to design a diode to operate at a particular forward voltage. Since every diode is different, this is a ridiculous goal, which cannot be achieved in practice. The point on the IV curve where the diode conducts a substantial current, like 10 mA, is not called the KNEE. It is just called the forward voltage. A KNEE occurs in a REVERSE BIASED Zener diode.
 

crutschow

Joined Mar 14, 2008
28,580
Since the diode current is a logarithm function of voltage, the apparent "knee", using linear scales, depends upon the current range of the test.
The simulation below, with a logarithmic horizontal current scale, gives a straight line, showing this log relation between voltage and current.
The slight curve at above about 10mA is due to the intrinsic ohmic resistance of the diode.

1635483975806.png
 

Papabravo

Joined Feb 24, 2006
17,646
You can get current as a function of voltage or voltage as a function of current and both representations should give you the same set of points. Straight lines on a semi-log grid imply an exponential relationship.
 
Top