- Why does the current increase so sharply after about 0.7V in a silicon diode?
- The exponential term in the Shockley equation dominates the current once the applied voltage V surpasses the thermal voltage (kT/q, ~26 mV at room temp) multiplied by the ideality factor. For silicon, the combination of the built-in potential and the exponential rise creates the common observation of a 'turn-on' voltage around 0.6-0.7V. This is not a fixed threshold but a region where the exponential function's value becomes very large very quickly.
- What does the ideality factor (η) represent, and why isn't it always 1?
- An ideality factor of 1 represents an ideal diode where current is dominated by pure diffusion of carriers in the neutral regions. In real diodes, factors like carrier recombination or generation within the depletion region contribute to the current. These processes have a different voltage dependence, requiring η to be between 1 and 2 (often ~1 for silicon, ~2 for high recombination) to fit the measured I-V curve to the Shockley model.
- How does temperature affect a diode?
- Temperature has two primary effects. First, it exponentially increases the reverse saturation current (I_0), as more electron-hole pairs are thermally generated. Second, it increases the thermal voltage (kT/q), which 'softens' the exponential curve. In forward bias, these competing effects mean that for a fixed forward voltage, the current increases with temperature, a critical consideration for thermal stability in circuits.
- Does the diode conduct in reverse bias?
- Yes, but very little. The Shockley equation predicts a small, essentially constant reverse saturation current (-I_0) for reverse biases larger than a few kT/q. This current is due to the minority carriers diffusing to the junction and being swept across by the built-in field. This model does not include reverse breakdown (Zener or avalanche), which occurs at much higher reverse voltages.