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There is no such thing as a 'voltage difference'! 'Voltage' means 'potential difference', so what you appear to be asking is "How do you get the largest potential difference difference?'! Potential difference is caused by the separation of charges between two points. The greater the amount of charge separation, the greater the potential difference.
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.
zener cut in voltage
Conventional LEDs are made from a variety of semiconductor materials, and radiate at different wavelengths all the way from infrared to ultraviolet. Their forward voltage drops range from about 1 volt to 6 volts, depending on the materials used. Any bias greater than the diode's forward voltage drop illuminates the device.
Unless you are using 'potential' in the general sense (i.e. "What is the possible voltage?"), there is no such engineering term as 'potential voltage'. Voltage is a synonym for 'potential difference', so your expression would then mean "What is the potential potential difference?"Do not mix up 'potential' with 'potential difference' (voltage); they are two different things.
Breakdown voltage is far greater than barrier potential. silicon:- break-down voltage :- 5v - 450 v barrier potential ;- 0.5v to 0.7 V
The potential barrier of a diode is caused by the movement of electrons to create holes. The electrons and holes create a potential barrier, but as this voltage will not supply current, it cannot be used as a voltage source.
One voltage is greater in thyristor whether forward breakover or reverse breakdown voltage. The greater of the two voltages in thyristor is forward breakover voltage.
It is stated that one of the voltage is greater in thyristor whether it be forward breakover or reverse breakdown voltage. It is also stated that the greater of the 2 voltages in thyristor is the forward breakover voltage.
forward breakover voltage is slightly smaller than reverse breakdown voltage
There is no such thing as a 'voltage difference'! 'Voltage' means 'potential difference', so what you appear to be asking is "How do you get the largest potential difference difference?'! Potential difference is caused by the separation of charges between two points. The greater the amount of charge separation, the greater the potential difference.
== 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.
In this circumstance, the clearance between components at different potential must be increased.
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.
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 normal diode with reverse voltage in excess of its rated breakdown voltage could fail. Resistance could be high (blown open) or low (shorted).
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.