Cutoff voltage is the point at which the battery is fully discharged. This is usually the point at which the device will shut itself off.
cut in voltage for si is 0.6 and for ge is 0.2
the maxmium voltage below which no current flows.....
An ideal diode would match the purpose of a diode without any of the drawbacks. The purpose of a diode is to control current flow - The diode "turns on" for current flowing in one direction, and "turns off" if current wants to flow in the other direction. Ideally, there would be no voltage drop across this diode when allowing current flow, thus no power loss. When the diode is "turned off" by a negative voltage, idealy there would be no current flow (no matter how large the negative voltage).
A A diode has two state either on [forward conduction .7 volts. Or reversed bias meaning the diode voltage polarity are revered or cut off, depends on the diode but it will leak some current , Anyhow that is reversed bias diode conditionForward conduction voltage can vary between about 0.6 volts and about 1.4 volts depending on the diode and the current. Also, some diodes, such as zener diodes, will conduct in the reverse direction. Reverse bias is simply where the anode is more negative than the cathode.
fet is a voltage controlled device...cut off voltage in fet refers to that voltage of the gate - source junction at which the current flow through channel is zero
If the current through a coil is interrupted, the coil generates a high voltage (such as in the old car ignition coils). The diode provides a path for the current to decay naturally, thus avoiding the high voltage.
If you're looking for a definition, it's: the voltage at which, a diode can be considered a "short circuit" or low-value resistor It varies with each diode, but most have approximately 0.6 or 0.7 Volts across them when you get almost 1mA flowing FORWARD through them. For light emitting diodes (LEDs), it varies between diodes and is largely dependent on the colour of the light. Green ones typically have 1.3V @ 1mA, red = 1.8V @ 1mA, and higher for other colours. Infrared LEDs usually have 1.1V @ 1mA. Higher cutoff voltages occur at higher forward currents, meaning that at 1mA, Vf might be 1.8V for a certain diode, but at 10mA Vf is maybe 1.9V. One important side note is that reverse current is still possible, but is so small it's usually negligible. Also, it's not recommended to force current backwards through a diode (exception: Zener diodes) because it usually requires a higher voltage to accomplish this. The cutoff voltage of a diode is the maximum voltage that the diode can withstand in the revers biase above which the device will be destroyed.
The cut in voltage is that voltage where after the current increase rapidly and it's value is different-2 for different type of semiconductor. for silicon it is 0.7 and for germanium it is 0.3 volt. It means if you made a diode from silicon than applied voltage below 0.7 volt will not able to flow current (if flow than it is in term of few micro ampere) . it means diode will stay in off mode and for germanium same things will happen and it will rest on off condition below 0.3 volt(applied).
It's a diode that is used to limit a voltage to some desired value. Importantly, it's used to limit a transient voltage or an alternative voltage. The simplest example is that diode placed across a relay's inductive coil. As current is cut off through the coil, a large induced voltage appears that can cause damage to the circuit. A diode can be placed in circuit to catch/damp the induced voltage.
An ideal diode would match the purpose of a diode without any of the drawbacks. The purpose of a diode is to control current flow - The diode "turns on" for current flowing in one direction, and "turns off" if current wants to flow in the other direction. Ideally, there would be no voltage drop across this diode when allowing current flow, thus no power loss. When the diode is "turned off" by a negative voltage, idealy there would be no current flow (no matter how large the negative voltage).
If the DC source biases the diode off, then the output will be zero. If it biases the diode on, then the output will be DC, with the voltage being nearly the same as the input voltage.
A A diode has two state either on [forward conduction .7 volts. Or reversed bias meaning the diode voltage polarity are revered or cut off, depends on the diode but it will leak some current , Anyhow that is reversed bias diode conditionForward conduction voltage can vary between about 0.6 volts and about 1.4 volts depending on the diode and the current. Also, some diodes, such as zener diodes, will conduct in the reverse direction. Reverse bias is simply where the anode is more negative than the cathode.
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fet is a voltage controlled device...cut off voltage in fet refers to that voltage of the gate - source junction at which the current flow through channel is zero
fet is a voltage controlled device...cut off voltage in fet refers to that voltage of the gate - source junction at which the current flow through channel is zero
I diode allows current to flow in only one direction. Therefore, if a lamp is "on" in a DC circuit, and the diode in series with the lamp is reversed, the light will be turnned off due to the diode blocking current flow (unless the voltage is above the breakdown voltage of the diode - if this is the case, the diode will fail). If this is an AC circuit, every half cycle the diode will turn on, then the next half cycle it will turn off. To your eye, the bulb will most likely appear slightly dim due to this on then off cycling. If the diode is reversed, there will be no apparent change. The difference is the half cycle the diode would have been off before reversing, it will now be on, etc.
A diode is voltage and current sensitive it follows an exponential curve and lower then .6v it is considered off and above .6v is considered on.
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.
No, a diode can rectify an AC signal but is not able to amplify an AC signal. Diodes are two layer devices whereas transistors have three. It is this very thin 'base' region in the transistor that gives it the ability to give a voltage or current gain.