Spice Model

[mady]

Hello Everyone,

I want the Spice Model for SR1660 which is a Schottky Barrier Rectifier. I have search in the Google but I couldn't found it.
If the Spice model is not available, how to write the Spice code for the model by seeing the datasheet?

Here, I have attached the datasheet of the SR1660 device.

 

[MikeMl]

This one, which is part of the standard.dio lib in the release, is probably close enough...

If not, copy out its .model parameters, and tweak the ones dont like...

.model MBRB2545CT D(Is=19.4n Rs=.01 N=.733 Eg=.718 Xti=.535 Cjo=2.05n Vj=.4 M=.41 Iave=25 Vpk=45 mfg=GI type=Schottky)

 

[mady]

Hello MikeMl,

I am not understanding single parameter. Please, let me know clear explanation. Otherwise, write the Spice code for my device which is SR1660.

 

[ronv]

Hello MikeMl,

I am not understanding single parameter. Please, let me know clear explanation. Otherwise, write the Spice code for my device which is SR1660.

Not sure how much this will help as many parameters are not in the data sheet. Easies way to come pretty close is to pick a diode from the standard library with the same current and voltage rating and family and use it.

https://www.acsu.buffalo.edu/~wie/applet/spice_pndiode/spice_diode_table.html

 

[MikeMl]

... Easies way to come pretty close is to pick a diode from the standard library with the same current and voltage rating and family and use it.

Which is what I did. That one is close enough for most sims...

 

[MikeMl]

Hello MikeMl, I am not understanding single parameter. Please, let me know clear explanation. Otherwise, write the Spice code for my device which is SR1660.

And what are you going to do?

 

[mady]

Actually, I have followed the below link:


In the above link, they are explained and calculated the parameter values of Silicon Diodes. But, In that explanation they are not clearly explained about some parameters which are VT, RS, TT, VJ, EG and ETI values. So, how these values can be calculated ?

And also said in that, For some parameter (i.e: RS, TT, VJ, EG and XTI) values for the Schottky Diodes consider as a default values and some others are consider near to the default value which are shown in table 2. So, how they are considered those values? Is there any formula's for that ?

In the table 2, 1N5711 SK and 1N5712 SK are the Schottky diodes. Now, we can see comparison between the default values and Schottky diode values.

In my case, I am using the Schottky Diode (SR1660). So, how I have to design those parameter (i.e: RS, TT, VJ, EG and XTI) values. Especially, how they are calculated VT value ?

Please, let me know clear explanation.

 

[ronv]

I don't see a Vt value anywhere in the tables.

 

[mady]

The VT value you can see under this equation (ID = IS(eVD/nVT -1)) which is below the table2.

 

[MrAl]

Hi,

You can get a pretty good approximation to Is with this:
Is=e^a

where a is:
a=-(Id1*log(Id2)*Vd3-log(Id1)*Id2*Vd3-Id1*log(Id3)*Vd2+log(Id1)*Id3*Vd2+Id2*log(Id3)*Vd1-log(Id2)*Id3*Vd1)/(Id2*Vd3-Id1*Vd3-Id3*Vd2+Id1*Vd2+Id3*Vd1-Id2*Vd1)

where log(x) is the natural logarithm ln(x), and
where the currents Id1, Id2, Id3 and voltages Vd1, Vd2, Vd3
are three measured points on the diode curve, with the lowest current at least maybe 5ma. Cant go too low on the current though or the approximation will move farther away from the true solution. The points should be spread apart somewhat too, with current also not too high. So maybe 5, 10, and 20 ma.

From that you can solve for the other constants N and Rs.

 

[mady]

Hi,

How to calculate the exact emission co-efficient value of Schottky Diode ?

I have followed the link which is mentioned below.

http://www.tietze-schenk.com/tsprben.pdf

In that pdf mentioned that (page.no: 16 under equation 1.6) the emission coefficient n = 1 for ideal diodes and for real diodes n = 1.....2, but they are not mentioned the exact value for Schottky diodes or real diodes. So, how we can choose the exact emission co-efficient value for Schottky diodes.

(or)

We can choose any one of the value which is between 1 to 2 ?

 

[MikeMl]

How accurately do you need to model that big diode?

Whatever you come up with should closely follow the I/V curve I showed you back in post #2.

 

[MrAl]

Hi,

How to calculate the exact emission co-efficient value of Schottky Diode ?

I have followed the link which is mentioned below.

http://www.tietze-schenk.com/tsprben.pdf

In that pdf mentioned that (page.no: 16 under equation 1.6) the emission coefficient n = 1 for ideal diodes and for real diodes n = 1.....2, but they are not mentioned the exact value for Schottky diodes or real diodes. So, how we can choose the exact emission co-efficient value for Schottky diodes.

(or)

We can choose any one of the value which is between 1 to 2 ?

Hello again,

In post #10 i gave you a formula for calculating Is as that's the hardest one to solve for. Once you have Is, you can insert that into the diode equation and solve for N and Rs, or just solve for N and Rs and insert Is later:

N=(Id1*Vd2-Id2*Vd1)/((Id1*log(Is+Id2)-Id2*log(Is+Id1)+Id2*log(Is)-Id1*log(Is))*Vt)

Rs=-(log(Is+Id1)*Vd2-log(Is)*Vd2-log(Is+Id2)*Vd1+log(Is)*Vd1)/(Id1*log(Is+Id2)-Id2*log(Is+Id1)+Id2*log(Is)-Id1*log(Is))]

These two provide the N and Rs constants. Note again log(x)=ln(x) here.
Points Id1,Vd1 and Id2,Vd2 can be any two points on the original curve, but since we needed three points to solve for Is we can use two of those points, or measure two more sets.

In the attachment, the blue plot is the original diode and the red plot is the diode curve obtained by extracting three point values from the original diode curve. You can probably get even more accurate if you are more careful with the measurements, but you can see how similar they are already.
Id is along the horizontal (0 to 1 amp) and Vd vertical (0 to 500mv).
The original diode was a Schottky type diode, where sometimes two diodes in parallel are used. But you can see they are pretty similar using just one diode (red curve) to model the two in parallel (blue curve).

 

[mady]

Hello MrAl,

How to measure the three point on the diode curve ? It's just like choosing the points on the I-V characteristic or Is there any methods to measure those three points ?

In order to calculate the Is, N and Rs values, in which I-V characteristics (Forward or Reverse) we have to assume as a three point ?

In my case, the datasheet showing that Forward & Reverse characteristics. So, I am bit confusion ?

If you consider to calculate from Forward I-V characteristics, there are not drawn below the 100A. So, can I choose those three points between 0.1A to 1A. you can see in the datasheet which is attached below.

I am considering Id1, Id2, Id3 = 0.1A, 0.5A, 1A and Vd1, Vd2, Vd3 = 0.34V, 0.39V, 0.44V from the forward characteristics (see at fig:4).
Is consideration, right?

 

[MrAl]

Hello again,

Well, that's a 16 amp diode so you should probably pick one point close to the higher current. Maybe even at 20 amps. So the forward current points might be at 0.1, 1.0, and 20.0 amps and whatever voltage you can read off the graph for those current values.
This is for the forward conduction mode which is usually the most important. You can fill in for the reverse mode with another diode arranged as a clamp, and a rather large parallel resistor.

Ideally what we would do is a curve fitting using a sum of squared errors, but that is much more involved where we need a program to calculate all of the required deviations and sums. In that case we would take a lot of measurements over the entire curve.

 

[mady]

Hello MrAl,

Link 1:
https://www.google.co.nz/search?tbo=p&tbm=bks&q=isbn:1439860467

In the above link saying that (page 203 last paragraph), The most important parameters for power electronics applications are IS, BV, IBV, TT, and CJO.

So, these values are enough to calculate in order to use as a diode (i.e rectifier) in power electronics applications. (or) I have to calculate all of them which is mentioned the in the below link.

Link 2:


Actually, I have calculated IS, CJO and N by following the above two links and the formula gave by MrAl.

But, there are not clearly mentioned in the above two links about the following questions……

1) How to calculate this Ohmic Resistance RS ? formula for RS; ( value = RS / area ); how to calculate value and area ?

2) How to calculate Transit time TT ? In the data sheet they are not mentioned TT value. In order to calculate TT, how to know Reverse recovery charge, QRR and IFMif those values which is not mentioned in the data sheet?

3) How to calculate grading coefficient M ? From the above 2nd link (in the 2nd table), M = 0.333 they are consider for Schottky Diodes but the default is 0.5. how they are consider 0.33 for Schottky Diode ?

4) How to calculate activation energy EG ? From the above 2nd link (in the 2nd table), EG = 0.69 they are consider for Schottky Diodes but the default is 1.11. how they are consider 0.69 for Schottky Diode ?

5) How to calculate IS temperature exponent XTI ? From the above 2nd link (in the 2nd table), XTI = 2 they are consider for Schottky Diodes but the default is 3. how they are consider 2 for Schottky Diode ?

6) How to calculate the Junction Potential VJ? From the above 2nd link (in the 2nd table), The VJ value some times they consider as 0.6 and 1. the default is 1. how they are consider 0.6 for Schottky Diode ?

So, If the parameters data not available in the data sheet, how I have to choose the values.
Can I consider as a default values (or) what ever the values are chosen for Schottky
Diodes which is mentioned in the table 2 from the above link 2.

Please, let me know the good explanation about all 6 questions ? then only I will get full confident about how to develop the spice model for Schottky diodes!

 

[alec_t]

Why do you need a model for that precise diode? Simulations are only approximations of reality after all, and for most sims any diode with similar properties (e.g recovery time, current rating) to the target one will be 'good enough'. Is there something special you are trying to simulate, or is this an academic exercise?

 

[MrAl]

Hello again,

Are you reading the previous posts? The method i told you about solves for Is, N, and Rs.
But assuming you want to find Rs a different way, if you run the diode at high current near it's rating the slope is mostly due to the resistance Rs.

To calculate TT without knowing anything about the dynamic characteristics of the diode is impossible. Since that poor data sheet doesnt show much, it's almost worthless. Find a better data sheet or switch to a diode that actually has a good data sheet.
The only other resort is to measure TT yourself by driving the diode with a square wave through a resistor, forcing it into forward conduction on the positive half cycle and into reverse on the negative half cycle, measuring the current on a good scope. You need to be able to observe the area where the current goes negative for a brief period of time. Measuring that area A and the peak negative current Ip you can estimate TT by:
TT=A/|Ip|

 

[mady]

In my Master thesis, the task is to design and evaluate the circuit in both simulations and hardware board (PCB). If, I use the same components in the simulation and hardware then only I will get the results which are approximately same. Then, only I will finish my task.

 

[mady]

Hello MrAl,

I have followed the two methods in order to get the value of IS.

Method 1:
Is=e^a
where a is:
a=-(Id1*log(Id2)*Vd3-log(Id1)*Id2*Vd3-Id1*log(Id3)*Vd2+log(Id1)*Id3*Vd2+Id2*log(Id3)*Vd1-log(Id2)*Id3*Vd1)/(Id2*Vd3-Id1*Vd3-Id3*Vd2+Id1*Vd2+Id3*Vd1-Id2*Vd1)
Answer is : IS = 5.6µA

Method 2:
ID = IS(e(VD/nVT )-1)
Here: ID, VD and VT values are determined from the datasheet and 'n' value only I have assumed as a '1'
This information got it from the below two links.


https://www.google.co.nz/search?tbo=p&tbm=bks&q=isbn:1439860467

Answer is : IS = 26pA

So, now I am confusion which method I have to follow that is why keep on rising up the questions.

Please, Can you tell me which Method is right? and Which I have to follow.