When a p-n junction is formed then holes and electrons combine near the junction and some portion near the junction neutralizes.Now more electrons and holes cannot combine because of the existence of the neutral layer.They need energy to cross that layer.This raises a resistance between the flow of electrons and holes.This gives rise to the barrier voltage.
A PN junction allows current to flow when it is forward-biased, meaning the positive terminal of a voltage source is connected to the p-type material and the negative terminal to the n-type material. This reduces the barrier potential at the junction, allowing charge carriers (holes and electrons) to recombine and flow across the junction. In contrast, when the junction is reverse-biased, the barrier potential increases, preventing current flow.
Cut in voltage is the minimum voltage required to overcome the barrier potential. In other words it is like trying to push a large boulder....it may not be possible to push a large boulder by one person but it may be done if 2 or more people try to push it together depending on the size of the boulder.....similarly....the charge carriers in the barrier region have a potential energy of about 0.6V for Silicon and about 0.2V for Germanium. so in order for the diode to conduct, it is required to overcome the potential of the charge carriers in the junction barrier region and hence only if a potential more than that of the barrier potential (cut off voltage) is applied, then electrons flow past the junction barrier and the diode conducts.
The potential across a pn junction is called potential barrier because majority charge carriers have to overcome this potential before crossing the junction.
when the p-n junction is heavily doped p-n junction diode has very sharp breakdown voltage.
The forward voltage at which the current through the junction starts increasing rapidly, is called the knee voltage or cut-in voltage.
The voltage across a forward-biased PN junction in a semiconductor diode or transistor.
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barrier voltage decreases by 2mV for every 1 degree increase in temperature.
No, we don not consider the barrier voltage of a diode to be able to act as a voltage source. The barrier voltage arises during construction of the p-n junction, and it results from charge separation. Separating charges results in voltage, but this difference of potential cannot be tapped as a voltage source because it cannot supply current the way we understand conventional voltage sources are able do.
A PN junction allows current to flow when it is forward-biased, meaning the positive terminal of a voltage source is connected to the p-type material and the negative terminal to the n-type material. This reduces the barrier potential at the junction, allowing charge carriers (holes and electrons) to recombine and flow across the junction. In contrast, when the junction is reverse-biased, the barrier potential increases, preventing current flow.
When sufficient forward voltage is applied across the junction, the electric field opposing the further diffusion of electrons from n-type to p-type semiconductor gets lost. The electric field created due to the application of the forward voltage opposes that of the barrier potential and finally vanishes the barrier completely.
A forward biased p-n junction is when an external voltage is applied in such a way that the positive terminal is connected to the p-type material and the negative terminal is connected to the n-type material. This reduces the barrier potential, allowing current to flow through the junction.
Whenever two dissimilar conductors touch a "potential barrier" forms. All conductive materials have a voltage above zero that an electron must have to enter the material. In true conductors, this voltage is very low. In semiconductors, it can vary, but is usually in the 0.25 to 6.8 Volt range. In insulators, it can be very, very high. When two semiconductors or a metal and a semiconductor touch the difference is polarity sensitive. The higher the "band gap", the voltage that must be overcome to enter the "conduction band", the higher the voltage drop in the forward bias direction. Think of the "band gap", or potential barrier, as being like a curb on a road and sidewalk. It's easy to ride a bike off the sidewalk over the curb onto the road, but not so easy the other way.
== When we make a semiconductor junction (a p-n junction), the electric fields force charges to shift creating what is called a depletion region. This depletion region forms a potential barrier across the junction. This potential barrier has a voltage associated with it, and that voltage is 0.3 volts (approximately) for germanium semiconductor material, and 0.7 volts (approximately) for silicon semiconductor. The terms we apply to this barrier potential are the built-in voltage (or potential), junction voltage (or potential), and contact potential. Use the link below to check facts and review some other closely related material.
The typical value of the barrier potential for a germanium diode is around 0.3 to 0.4 volts. This barrier potential is the voltage required to overcome the potential barrier at the junction of the diode and allow current flow in the forward direction.
Cut in voltage is the minimum voltage required to overcome the barrier potential. In other words it is like trying to push a large boulder....it may not be possible to push a large boulder by one person but it may be done if 2 or more people try to push it together depending on the size of the boulder.....similarly....the charge carriers in the barrier region have a potential energy of about 0.6V for Silicon and about 0.2V for Germanium. so in order for the diode to conduct, it is required to overcome the potential of the charge carriers in the junction barrier region and hence only if a potential more than that of the barrier potential (cut off voltage) is applied, then electrons flow past the junction barrier and the diode conducts.
cut in voltage *** for silicon is 0.7volts and that for germanium is 0.3volts.According to Millman and Taub, "Pulse, Digital and Switching Waveforms", McGraw-Hill 1965, the cutin (or offset, break-point or threshold) voltage for a silicon diode is 0.6, and 0.2 for germanium.Breakdown voltage is another thing entirely. It is the reverse voltage at which the junction will break down.