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I have an LED that I'm working with, actually it's not a discrete part but on silicon. Anyway all I have is a curve for it and I was trying to work up a spice model in LT-spice. I followed this question and tried two methods from there with Martyn's python method giving me the best results.

For LT spice I took the three parameters he puts out and made a model statement: .model led D ****DC Model Parameter*** +is=1.87E-17 n=9.275 +rs=5500

Which gives me a curve that is close to my original data points at low voltages but diverges at higher voltages like so: enter image description here

Orange is the measured data and blue is the model.

I've tried adjusting the parameters manually for a few hours but I can't seem to get something that matches the higher end. I feel like either I'm missing something or maybe I need to add another element other than just a diode? I seem to remember opening some professional models and seeing a few components inside.

How can I make my model match my actual datapoints better?

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You mentioned professional models with a few components inside. A full device model certainly may include some AC components to deal with charge storage and inductance. But you are developing just the DC model. So don't worry about those details here.

The DC equation is a modified Shockley equation that looks something like this:

$$V=n V_T \operatorname{ln}\left(\cfrac{I}{I_S}+1\right)+I\cdot R_S$$

You want to figure out \$n\$, \$I_S\$, and \$R_S\$.


As \$V_T=\frac{k T}{q}\$ and is temperature-dependent, you do not want to complicate your life by making temperature yet another variable. So make sure that you control the diode's die temperature and keep it a constant. The best way to do this is to:

  1. Allow the LED temperature time to equilibrate with the ambient temperature -- so if you are holding the LED for a while, give it time to cool down to room temperature, for example,
  2. avoid making measurements over long periods of time where the ambient temperature has time to move,
  3. make your measurements quickly to avoid heating the LED die,
  4. avoid using higher currents that are likely to cause significant heating of the LED's die and, when necessary (it will be), use only very short pulse currents with long delays between (but not so long as to violate (1) or (2) above.)

(Or you can use an elevated, stable temperature heater to keep the LED at some fixed temperature.)


Let's look at the overall resistance illustrated by the above equation. This is found by computing \$\frac{\textrm{d} V}{\textrm{d} I}\$ of a slightly simplified version (ignoring the +1 term for now):

$$R \approx \frac{n V_T}{I} + R_S $$

That shows us that there are two parts to the resistance. And resistance is the slope of the \$I\$ vs \$V\$ curve. This breaks the curve into two parts, with a transition region between them. The two regions are where \$I \ll \frac{n V_T}{R_S}\$ and where \$I \gg \frac{n V_T}{R_S}\$.

There is a third region which can't be seen from this equation, but is present in the fuller equation presented earlier above. This is where \$I \approx I_S\$. You will need to stay away from this area because it is near here that the ignored +1 term matters more and complicates your life.

So you need to make at least two different measurements in at least two different regions. One where \$I_S\ll I \ll\frac{n V_T}{R_S}\$ (I'll call this Region II) and the other where \$I \gg\frac{n V_T}{R_S}\$ (Region III.) Region I is where the +1 term matters. You should stay away from Region I, entirely.

Region II may not heat up the LED die that much (good.) But Region III likely will, so you probably need pulse measurements for that region.

For Region II, you are attempting to figure out the value of \$n\$ and \$I_S\$. If you plot Region II measurements on a \$y=\operatorname{ln}\left(I\right)\$ vs \$x=V\$ curve, you will find that they fall on a fairly linear line. If not, you aren't fully within the region. Region I will curve away from the line at very low currents and Region III will also curve away from the line at very high currents. So you need to find and then make your Region II measurements in the linear area between Region I and Region III.

(There are transitions between these three Regions, too. Stay away from those transition areas, as well.)

Once you have these Region II measurements, \$n\$ is then inversely proportional to the slope of the plotted line. \$b=\operatorname{ln}\left(I_S\right)\$ is the y-axis intercept. If you make more than two such measurements at different values of \$I\$, then a standard linear least squares fitting routine can compute the slope (\$m\$) and intercept (\$b\$) for you. The slope you get from this will be \$m=\frac{1}{n V_T}\$, so just solve that: \$n=\frac{1}{m\cdot V_T}\$. The intercept will be in terms of \$\operatorname{ln}\left(I\right)\$, so you need to raise \$e\$ to that power to get \$I_S\$. Such fitting algorithms are everywhere. So you shouldn't have any trouble with that. (If you only make two measurements you can use simple algebra to figure out the resulting slope and intercept.)

To get \$R_S\$, you need to use large, pulsed currents in Region III, with those long delays between. Just two such measurements should be enough. (Again, to figure out where Region III is, you need to take a variety of pulsed measurements to probe that region and find out where it lays. Pick currents that require voltages well past the knee, here.)

Again, take at least two good measurements in Region III. More is okay, but then you are back to that least squares fitting routine again. From these you can figure out the slope. And the slope is the value of \$R_S\$.


That process should get you what you need. It uses relatively simple math tools (simple algebra for two measurements in each region, or linear least square fitting if more than two measurements in each region), too.

One thing I can see from your curve is that I don't think you have enough data to figure out \$R_S\$ well. You'll likely need higher current measurements.


To make the above concrete, imagine that \$V_T=26\:\textrm{mV}\$, \$n=3\$, \$I_S=1\:\textrm{pA}\$, and \$R_S=2\:\Omega\$. Region II is then \$1\:\textrm{pA}\ll I \ll 39\:\textrm{mA}\$ and Region III is \$I \gg 39\:\textrm{mA}\$.

Probe Region II at \$I_1=500\:\mu\textrm{A}\$ and \$I_2=3\:\textrm{mA}\$ (I picked those to create some interesting errors) to get \$V_1\approx 1.563\:\textrm{V}\$ and \$V_2\approx 1.708\:\textrm{V}\$. (I'm assuming we can only measure to four places.) On the log-lin plot, the slope is \$m=12.3569619\$ (It is valid to retain fuller precision for m [and for b, later].) Therefore, we compute \$n=\frac{1}{m V_T}\approx 3.1\$ -- which is close given the mV precision of the voltage measurement. The intercept is \$b=-26.9148338\$ and this leads to \$I_S=e^b\approx 2.05\:\textrm{pA}\$. This appears to be a more substantial error. But it is due to the effect of \$R_S\$, which at \$3\:\textrm{mA}\$ drops about \$6\:\textrm{mV}\$. Had I chosen a smaller high end current, such as \$1\:\textrm{mA}\$, I would have gotten about \$1.4\:\textrm{pA}\$ instead, which is closer.

You could then probe Region III with \$I_3=60\:\textrm{mA}\$ and \$I_4=100\:\textrm{mA}\$ (for very short periods) to get \$V_3=2.056\:\textrm{V}\$ and \$V_3=2.176\:\textrm{V}\$. (Again, these two currents are chosen close to one side of Region III to create interesting error, as well as to be practical instead of impractical.) From that, it is easy to find \$R_S\approx 3\:\Omega\$. Here, it's also wrong but not too far off. But you also need to take note that the dynamic part of the resistance was significant at \$1.34\:\Omega\$ and \$0.81\:\Omega\$, respectively. (Assuming our Region II parameter estimates.) So, of course there's an error. You could use that prior "knowledge" from Region II, namely our belief that \$n\approx 3.1\$ and \$I_S\approx 2.05\:\textrm{pA}\$, to then subtract about \$1\:\Omega\$ (a rough average of the two computed dynamic resistance values) to find \$R_S=2\:\Omega\$, which is of course much better.


All this should give you some ideas about how to figure things out and it manages to avoid overly complex mathematics by focusing on the dominant issues at hand in each region.

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When you plot Log(Id) vs. Vd and do not get a straight line, this means it's never going to be fit well by the standard model. In general you should always extract exponential modeled data (diodes or transistors) using Log(I) data because this means you'll be fitting a straight line (which is amenable to linear regression).

So first: are you using Log(Id) data to do the parameter extraction using linear regression to obtain IS and N? If not do this first!!

If that still doesn't work, the short answer is the diode model in SPICE presumes the diode is "ideal" (classic diode model equation) so when it's not, it won't fit and there's nothing you can do with the base model to improve that.

The standard fixes for this are 1) choose a corner most relevant to your particular application that needs to fit best and only worry about fitting that corner (e.g. make certain parameters that dominate that operation regime more accurate), or 2) use a macro-model using subcircuits to hack your way to a better fitting model. Each has issues.

This is BTW a super common issue with SPICE - the models are nothing like real devices most of the time so you have this issue. This is part of why the profession of "SPICE model extraction" can never be made "turn-key" run by monkeys. For example, pretty much all the semiconductor models are based on 1-dimensionally uniform devices but every real device has side-walls that have active PN junctions or doping. You have to think about the device manufacture and device physics to know how to model that.

For economically important technologies like MOSFETs companies can afford to research and invent new models that fit the current technology better. For most everything else we are still using 1970 vintage device models which are baked into SPICE. Since pretty much EVERY variant of SPICE (PSPICE, HSPICE, LtSPICE, etc.) is no more than 1%-3% deviation from the original SPICE, so the issue persists.

Here's a paper on LED modeling for better fits that might be helpful if you want to go the "macro model" route:

http://www.keysight.com/upload/cmc_upload/All/Diode.pdf

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Your numbers are way off from what are seen in typical III-V LEDs. The rs is very high, for one. And n is more often in the 2-3 range. But if these are silicon LEDs, that's a pretty odd beast, and anything might be possible.

You might want to talk to whoever designed the device (I'm assuming it's a research device, not something bought off the shelf) and see what they recommend.

Often when modeling odd devices like this, finding out what the parasitics are is critical to getting a good model.

For example, you might need to include a resistor in parallel with the main diode to model a leakage path. Or a second (Schottky) diode in series to represent the metal-semiconductor junction at the contact to your LED.

As jonk says, if you stick to the simple parasitics allowed in the SPICE diode model, the rs term will give the slope at voltages well above turn-on. So you should match it to the slope in your measurement at the maximum voltage measured. Then keep that fixed while you adjust the other parameters to try to get a match to the rest of the curve.

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I assume that the blue curve is your actual LED device and orange is trying to predict its characteristic. The blue device is clearly an LED in series with a resistor so your model needs to be put in series with the model of a resistor.

Here's a picture of the forward characteristic of an LED with resistance added in series: -

enter image description here

As you can see, adding the resistor linearises the response above the basic point where forward conduction begins. Picture stolen from here.

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  • \$\begingroup\$ Yes, the model statement "RS" accommodates a series-resistor that SPICE calls "parasitic resistance". Its default is zero. The OP graph suggests it should be closer to 6000 ohms. \$\endgroup\$ – glen_geek Dec 29 '16 at 15:32
  • \$\begingroup\$ @glen_geek I reckon it's more like 50 ohms - a 2 volt change takes current from 80 mA to 120 mA. \$\endgroup\$ – Andy aka Dec 29 '16 at 15:35
  • \$\begingroup\$ Am guessing the vertical axis labeled Curent is actually Current , since it has units of milliamps. Seems a pretty weak LED, even for high-efficiency...unless those units are wrong too. \$\endgroup\$ – glen_geek Dec 29 '16 at 15:42
  • \$\begingroup\$ @glen_geek hehe I misread those numbers - I think you are right, for a voltage change of 2 volts, current changes 0.4 mA =~5k. \$\endgroup\$ – Andy aka Dec 29 '16 at 15:46
  • \$\begingroup\$ Hi thanks for responding, actually orange is the measured data and blue is my model. It is a weak LED in the sense that it is only a few microns in diameter :) \$\endgroup\$ – confused Dec 29 '16 at 16:00

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