The I-V characteristics of an LED is:

We recall that an LED (Light Emitting Diode) is a $$p$$-$$n$$ junction device that starts conducting only when it is forward-biased and the applied forward voltage exceeds a certain critical value called the cut-in voltage or threshold voltage. For most visible LEDs this threshold voltage lies roughly between $$1.6\text{ V}$$ and $$3.0\text{ V}$$ depending on the semiconductor material and the colour.

First we analyse the forward-bias region. When the applied forward voltage $$V_F$$ is less than the threshold value $$V_T$$, the current $$I$$ is extremely small because the potential barrier inside the junction is not yet overcome. Mathematically, the diode current in forward bias is governed by the diode equation

$$I = I_S\!\left(e^{\dfrac{qV_F}{kT}} - 1\right),$$

where $$I_S$$ is the reverse saturation current, $$q$$ is the electronic charge, $$k$$ is Boltzmann’s constant and $$T$$ is the absolute temperature. For $$V_F \lt V_T$$ the exponential term is still close to $$1$$, so $$I \approx 0$$. The graph therefore hugs the current axis near the origin.

Now, the moment $$V_F$$ reaches and slightly exceeds $$V_T$$, the exponential term $$e^{qV_F/kT}$$ grows very rapidly. Consequently $$I$$ increases almost exponentially with a very small additional increase in $$V_F$$. On an I-V graph this appears as a sharply rising curve beyond the knee point (the knee is located at $$V_T$$).

Next we consider the reverse-bias region. When the LED is reverse-biased (i.e. the $$p$$ side is connected to the negative terminal of the supply and the $$n$$ side to the positive), the junction potential barrier widens. The current now reduces to the tiny reverse saturation current $$I_S$$. For practical purposes this current is so small that it almost coincides with the horizontal voltage axis. Only if the reverse voltage is made extremely large (far beyond normal operating conditions) will breakdown occur, but LEDs are never operated in that region. Hence, in the region $$V \lt 0$$, the current is essentially zero and the curve nearly coincides with the voltage axis.

Summarising the above two behaviours, the complete I-V plot of an LED must:

1. Almost lie on the voltage axis for $$V \lt 0$$ (reverse-bias) showing negligible current.
2. Remain nearly on the current axis for small positive voltages $$0 \lt V_F \lt V_T$$.
3. Exhibit a pronounced knee at $$V = V_T$$.
4. Rise steeply and non-linearly (exponentially) for $$V \gt V_T$$.

Among the four given schematic options, only Option (3) displays a curve which:

$$ \text{(i) lies very close to the }V\text{-axis for }V \lt 0,\\[2pt] \text{(ii) stays close to the }I\text{-axis until }V = V_T,\\[2pt] \text{(iii) rises sharply after the knee.} $$

The other options either show a linear region where the rise should be exponential, or depict significant reverse current which does not match the actual LED behaviour.

Therefore, by matching the theoretical I-V characteristics with the sketched options, we conclude that Option (3) is the only correct representation.

Hence, the correct answer is Option 3.