Reverse breakdown voltage setting for the diodes in a solar cell model in SPICE

diodesltspicespice

I want to simulate the reverse breakdown behavior of a two diode model solar cell in SPICE.
The diodes configurations without considering the breakdown voltage are as below:

.model diode d( Is=1.5e-10,  xti=3)
.model diode2 d(Is=1.77e-6, n=2, xti=3)

With the given diodes configurations, I can simulate all characteristics before the breakdown voltage point.
I-V curve without adding reverse breakdown voltage code
By adding the breakdown voltage point to the diode model, I cannot see the curve in forward bias anymore and it does not work properly.

.model diode d( Is=1.5e-10,  xti=3, vrev=15, BV=15)
.model diode2 d(Is=1.77e-6, n=2, xti=3, vrev=15, BV=15)

enter image description here

In short, when I add the code for reverse breakdown voltage point, my diode does not work properly in forward bias! Please help me with this problem.

Best Answer

From LTWiki:

There are two types of diodes available. One is a conduction region-wise linear model that yields a computationally light weight representation of an idealized diode.

...

This idealized model is used if any of Ron, Roff, Vfwd, Vrev or Rrev is specified in the model.

So when you included the VREV parameter, you changed your model to use the piecewise linear diode model instead of the usual nonlinear model. Your other parameters (IS, XTI, and BV) are ignored and the piecewise linear model parameters that you didn't specify take their default values.. This could make a substantial change to the operating behavior of the diodes in your model.

Related Solutions

LED – How to Model an LED with SPICE

As you stated, there are 3 parameters that dictate the DC response of a diode. Those are the saturation current (IS), the emission coefficient (N), and the ohmic resistance (RS). I was able to fit the curve with a fairly high accuracy, so I'll document my model procedure.

The SPICE model for the diode closely matches the Schokley diode equation:

If = IS(e^(Vf/(N*Vt)) - 1)

where Vt = kT/q = 26mV at room temperature.

Get actual values from the graphs provided in the datasheet to use for comparison. The more points the better, and the more accurate the better. Below is a table that I estimated from the figure you provided:
```
Vf  If (mA)
1.3 0.001
1.4 0.010
1.5 0.080
1.6 0.700
1.7 5.000
1.8 20.000
1.9 40.000
2.0 65.000
2.1 80.000
```
Plug the values into Excel, and change the y-axis to a log scale. You should get a graph that looks identical to the original graph from the datasheet. Add another column for your graph, with If calculated from the forward voltage and the constants IS and N. We can use this configuration to iteratively find IS and N.
Solve for IS and N. We are trying to match the linear part of the graph (1.3 <= Vf <= 1.7). Adjusting IS will move the curve in the y-axis. Get the calculated graph to the same order of magnitude. The next step is to find the emission coefficient (N). N affects both the amplitude and the slope, so some adjustment of IS may be necessary to keep the curve in the same ballpark. Once the slopes match (the lines are parallel), trim IS so that the calculated data matches the datasheet values. I got IS = 1e-18, and N=1.8 for the diode you listed.
Identify RS. This is a bit tricky. RS is responsible for the curving of the current from 1.7V and above. Consider modeling the ohmic resistance as a resistor in series with the diode. As the current through the diode increases, the voltage drop across the ohmic resistance causes the forward diode voltage Vf to increase slower. At small currents, this effect is negligible.

The first thing to do is to get a ballpark estimate of RS to use in the more accurate solutions. You can calculate the effective value of RS from the datasheet values by back-calculating for Vf using the measured If. The voltage difference between the input value and the calculated Vf can be used with the forward current to generate a resistance. At the higher currents, this will be a good starting value.

To plot the diode current using RS, you need to first calculate the diode Vf given a voltage for the resistor-diode series combination. Wikipedia lists an iterative function - it converges easily if the resistor voltage drop is significant. This function was easy enough to set up in Excel. For Vf values below 1.8, I hard-coded the input value because the iterative function did not converge. Then take this Vf value to calculate the If of the ideal diode. I plotted this with the original datasheet graph.

Using trial and error, you should be able to get a RS value that gets pretty good overlap with the datasheet values. All that's left is to throw the model together in SPICE to verify your work.

Below is my diode model that I verified using HSPICE. The simulation data is almost a perfect overlay for the datasheet graph.

.model Dled_test D (IS=1a RS=3.3 N=1.8)

I used this article, which helped a lot with the diode spice parameters.

I cleaned up my spreadsheet, and tyblu has made it available for download here. Use at your own risk, results not guaranteed, etc... etc...

Electronic – Rationale for operating the diode in the (reverse biased) breakdown region

Three reasons:

First, operating in the forward orientation only allows operation at a single voltage, nominally about 0.7 volts for a silicon diode. Diode construction can be tailored to produce a wide range of breakdown voltages, with a consequent choice of different regulator outputs.

Second, your V-I curve overstates the sharpness of a forward-biased junction. There is no relatively flat portion other than in the vicinity of zero, and that's not very useful.

Third, with an exponential V-I curve, the forward-biased junction cannot be operated at useful current levels with good regulation.

Best Answer

Related Solutions

LED – How to Model an LED with SPICE

Electronic – Rationale for operating the diode in the (reverse biased) breakdown region

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