Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET
Below the knee (about 0.7 V for Si, 0.3 V for Ge) the applied forward voltage is not yet enough to fully overcome the built-in potential barrier of the depletion region, so very few carriers cross and the current is tiny. Once the applied voltage exceeds this barrier, the depletion region becomes very thin and carriers cross easily, so the current rises steeply. That is why the forward part looks like a flat line that bends sharply upward at the knee.
In forward bias the current is carried by majority carriers (electrons in n-side, holes in p-side), which are very large in number, so the current is large (mA). In reverse bias the junction blocks majority carriers; only the few minority carriers cross, giving a very small reverse saturation current in microamperes (uA). This huge difference in scale is why the two halves of the graph use different current units.
No. A resistor gives a straight line through the origin because I is proportional to V (constant resistance). A diode is non-ohmic: its V-I graph is curved and not symmetric. The current depends on the sign of the voltage (conducts in forward, blocks in reverse), so a diode does not obey Ohm's law. NCERT specifically contrasts the straight resistor line with the bent diode curve.
Reverse saturation current is the small leakage current that flows in reverse bias due to minority carriers being swept across the junction. It is called 'saturation' because increasing the reverse voltage does not increase it much: there are only a limited number of minority carriers available to cross, so the current stays nearly constant (a flat line) until breakdown.
Heating breaks more covalent bonds and creates more electron-hole pairs, so more minority carriers are available. This increases the reverse saturation current and also changes the forward behaviour, so heating affects the overall V-I characteristics of the diode, not just one side. This exact idea was tested in NEET 2018.
A normal p-n junction diode characteristic uses the first quadrant (forward: +V, +I) and the third quadrant (reverse: -V, small -I). A solar cell is special: because it generates its own voltage from light, its I-V characteristic lies in the fourth quadrant. NEET 2024 tested this exact point.
In a p-n junction diode, change in temperature due to heating:
Consider the following statements A and B and identify the correct answer: A. For a solar cell, the I-V characteristic lies in the IV (fourth) quadrant of the given graph. B. In a reverse biased p-n junction diode, the current measured in microamperes is due to majority charge carriers.
Two statements are given below: A. When the forward bias voltage across a p-n junction diode increases above a certain threshold voltage, the diode current increases significantly. B. This current is called reverse saturation current. Choose the correct answer.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
They are a graph of diode current I versus applied voltage V. In forward bias the current stays near zero until the knee voltage (about 0.7 V for Si, 0.3 V for Ge), then rises steeply. In reverse bias only a small, nearly constant reverse saturation current flows until breakdown.
It is the forward voltage at which the current suddenly starts rising sharply. It is about 0.3 V for germanium and about 0.7 V for silicon. Below this the diode barely conducts; above it the diode conducts freely.
Because its V-I graph is curved and depends on the sign of the applied voltage. It conducts strongly in forward bias and blocks current in reverse bias, so current is not simply proportional to voltage as it is in a resistor.
When the reverse voltage is increased beyond a critical value, the reverse current suddenly rises very steeply (a near-vertical line). This is the breakdown region, which special diodes like Zener diodes are built to use safely.
Forward current is large (milliamperes) because majority carriers flow, while reverse current is tiny (microamperes) because only minority carriers flow. Using different scales lets both parts be seen clearly on one graph.
