How to measure the saturation current of a diode

There's a few things that you can do:

1. Find a SPICE model for your particular LED. SPICE uses the Shockley diode model, so you can just pull Is and N from the model card.
2. Find a datasheet for the LED in question (with a graph indicating a voltage-current relationship) and model only Is and N, which is actually not very difficult. See my answer on "How to model a LED" for most of what you need to get started. Only solving for Is and N will work well until you start trying to use the equation for peak currents (high current, low duty cycle).
3. Take the forward threshold voltage of the diode, and solve the diode equation for Is. Guess about 1.8-2.0 for N and just be satisfied with the result. It won't track reality very well, but it will behave like a diode.

4. Take the values for this red LED and hope that they mostly work for your LED.

   IS=1a RS=3.3 N=1.8

Keep in mind that different color LEDs have different forward voltage drops, and that even diodes with the same color can have different voltages. Also, the diode equation neglects series resistance, which can have a significant effect at high currents. It is up to you to understand your accuracy requirements, and to know the limitations of your calculations.

Shockly diode equation is given by: $$\style{}{I=I_S(e^{V_D/nV_T}-1)}\tag1$$

Where,

• \$I_S\$ is the reverse bias saturation current.
• \$V_D\$ is the voltage across the diode
• \$V_T\$ is the thermal voltage
• \$n\$ is the ideality factor, also known as the quality factor or emission coefficient.

I see two material dependent parameters here:

1. The reverse saturation current, \$I_S\$ depends on the material.
2. The parameter \$n\$ depends on fabrication process and semiconductor material.

EDIT
From equation (1), $$V_D = n \cdot V_T \ln\left(\frac{I}{I_S}+1\right) \approx n \cdot V_T\ln\left(\frac{I}{I_S}\right)$$ $$V_D \approx n \cdot V_T \cdot \ln10 \cdot \log_{10}\left(\frac{I}{I_S}\right)$$

Assuming room temperature and \$n=1\$, $$V_D \approx 0.05916 \cdot \log_{10}\left(\frac{I}{I_S}\right)\tag2$$

Typical values of the saturation current at room temperature are:

• \$I_S = 10^{-12}\$ for silicon diodes;
• \$I_S = 10^{-6}\$ for germanium diodes.

For a current of 1.0 mA:

• \$V_D \approx 0.53 V\$ for silicon diodes (9 orders of magnitude)
• \$V_D \approx 0.18 V\$ for germanium diodes (3 orders of magnitude)

For a current of 100 mA:

• \$V_D \approx 0.65 V\$ for silicon diodes (11 orders of magnitude)
• \$V_D \approx 0.30 V\$ for germanium diodes (5 orders of magnitude)

Further increase in current won't cause much increase in \$V_D\$ (0.05916V per decade is the rate of change of \$V_D\$ with respect to \$I\$). Hence in the common cases (current in the range of mA), the voltage drop remains constant around 0.6V for silicon diodes.

And hence values of 0.6 or 0.7 Volts are commonly used as voltage drop for silicon diodes and 0.3 for germanium diodes.

source: wikipedia