Explain what happens to the charge carriers in a semiconductor light dependant resistor when it is illuminated.
Suggest why shining light onto an LED might be expected to generate a small voltage between its terminals.
Thx
A semiconducting light-dependent resistor excites more free electrons into the conduction band when light is present. In solid state physics language, the new photoelectrons are "raised above the Fermi level" of the semiconductor.
An LED is reversible device. Electrical potential between the electrodes liberates photons (light), but photons incident upon the junction can reverse the electron excitation process, driving electrons in the opposite direction and creating a voltage at the terminals.
To understand what happens to the charge carriers in a semiconductor light-dependent resistor (LDR) when it is illuminated, we need to know a bit about the band structure of semiconductors and how they conduct electricity.
In a semiconductor, there are two main energy bands that play a crucial role in its electrical behavior: the valence band and the conduction band. The valence band is the band where electrons are tightly bound to their respective atoms. The conduction band is the band above the valence band, and electrons in this band are free to move and conduct electric current.
In the dark, a light-dependent resistor is typically in a high-resistance state. This is because the energy gap between the valence band and the conduction band is relatively large, and only a few electrons have enough energy to cross this gap and become free charge carriers (conduction electrons).
However, when light is incident on the LDR, photons with enough energy can be absorbed by the semiconductor material. This absorption process transfers energy to electrons in the valence band, allowing them to surpass the energy gap and move into the conduction band. These newly excited electrons are now free charge carriers or conduction electrons.
As a result, the number of conduction electrons in the LDR increases, which leads to a decrease in resistance. This phenomenon is known as photoconductivity. The greater the intensity of light incident on the LDR, the more electrons are excited, and the lower the resistance becomes.
Now, let's consider why shining light onto an LED (Light-Emitting Diode) might generate a small voltage between its terminals. An LED is a device that emits light when a forward voltage is applied across its terminals. However, LEDs also exhibit the photovoltaic effect, which is the generation of a voltage when light falls on the junction of the LED.
When light photons with sufficient energy reach the junction of an LED, they can interact with the semiconductor material, exciting electrons across the energy gap. This creates a flow of electrons and holes (electron vacancies) within the material. These excited charge carriers can then separate under the influence of the built-in electric field of the junction.
The separated charges result in a voltage difference between the two terminals of the LED, creating a small photovoltaic effect. The magnitude of this voltage depends on various factors, such as the intensity of the incident light and the properties of the semiconductor material.
It's important to note that the photovoltaic effect in LEDs is relatively weak compared to dedicated solar cells or photodiodes, as LEDs are primarily designed for light emission rather than efficient power generation.
So, in summary, when light illuminates a semiconductor LDR, it excites more electrons into the conduction band, reducing its resistance. In the case of an LED, shining light on its junction can create a voltage difference between its terminals due to the photovoltaic effect.