Difference in real diode characteristics vs shockley equation

Question

I was wondering about the voltage drop in real diodes (0.7 V in Si diodes, 0.3 in Ge diodes etc.). According to my experience this drop is linked to the used material, but the shockley equation which should describe a diode never uses any material constant. So, how is the material constant introduced in a diode equation for modelling the additional voltage drop compared to an ideal diode?

Answer 1

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

Answer 2

The Shockley equation does NOT describe the equation of the part called "diode". It establishes the voltage-current relation of the pn junction only. This exponential function is linearized (for increasing currents) caused by the ohmic path resistance.

EDIT: The known exponential function has a factor Io (reverse saturation current) which is different for Si and Ge. This factor contains the carrier concentration factor which is proportional to the material-dependent potential gap (1.1 for Si and app. 0.6 for Ge).