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?
Real Diode Characteristics vs Shockley Equation – Differences Explained
diodes
Related Topic
- Voltage – Understanding Silicon Diode Threshold Voltage 0.7V
- Electronic – Characteristics of a Laser Diode Below Threshold
- Electronic – Charging an ideal capacitor through an ideal diode, forever
- Understanding the Temperature Coefficient of Diodes
- Shockley Diode Equation – Does it Apply to Schottky and Non-Silicon Diodes?
Best Answer
Shockly diode equation is given by: $$\style{}{I=I_S(e^{V_D/nV_T}-1)}\tag1$$
Where,
I see two material dependent parameters here:
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:
For a current of 1.0 mA:
For a current of 100 mA:
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