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.
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.
When the polarity of the battery is such that electrons are allowed to flow through the diode,then the diode is said to be forward-biased. Conversely, when the battery is "backward" and the diode blocks current, then the diode is said to be reverse-biased. A diode may be thought of as like a switch: "closed" when forward-biased and "open" when reverse-biased.
The difference in the 1N4007 diode and the 1N4007S diode is the voltage. The 1N4007S has a higher voltage but the meaning of the S is not listed.
Normally too higher voltage burns the diode.
when diode is supplied with a voltage higher than RIV in reverse bias, the diode will burn out and will have zero resistance.
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.
Forward biase the given diode by using a Variable resistor in the circuit. By adjusting the value of variable resistor you will adjust the voltage being applied to junction diode. First adjust the resistance such that no(negligble) current flows through the circuit. Now start decreasing the value of resistance. Note the voltage across resistor(Vr) when current just starts flowing through the circuit. Then Potential barrier of diode will be: Vb=V-Vr Vb:Barrier Potential V:Battery Voltage Vr:Voltage Drop across resistance when current just starts flowing through the circuit.
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
The barrier voltage of a diode is 0.7v for silicon and 0.3 for germanium. after this voltage is reached the current starts increasing rapidly... till this voltage is reached the current increases in very small steps...
The voltage across a forward-biased PN junction in a semiconductor diode or transistor.
ginago
zener diode
Potential barrier of silicon is 0.7, whereas potential barrier of germanium is 0.3
The contact potential of a pn-junction diode is signified by the turn-on or barrier voltage, which is the voltage beyond which non-negligible current con be measured flowing in the forward-bias direction. To put it simply, one can run a variable potential difference across a diode in it forward-bias direction until one measures a current. That is the contact potential of the diode. Theoretically, the contact potential is a function of the temperature and doping concentration, and intrinsic hole-electron pair concentrations. However, in the real world, there maybe other factors that will affect the contact potential of a diode.
if the diode is forward biasedwell practically the current flows in a circuit if and only if an effective resistance is present in the circuit, if we consider the diode to be ideal (barrier potential but no internal resistance) in this case an external resistance is required if we use the approximate model (both barrier potential and internal resistance are considered) we need not use an external resistance the internal resistance itself acts as the effective resistance.if the diode is reverse biased:-the same explanation applies even if the diode is reverse biased but one must take care that the reverse voltage drop on diode should not increase the peak inverse voltage mark the diode would be burnt or damaged if this phenomena occurs.So this can be prevented by adding suitable resistance to the circuit through which the voltage drop on diode can be managed
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.