forward bias 0 volts, reverse bias infinity volts.
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).
zener diode :zener diode operates under reverse bias voltageideal diode :ideal diode operates under forward bias voltage
forward drop is the same as any other silicon diode, about 0.7V
The forward biased voltage drop of a diode depends on the type of diode and the current through the diode. A typical silicon diode will exhibit a voltage drop between 0.6v and 1.4v depending on current. An LED might range from 2v to 3v. A germanium diode might go a low as 0.2v. Bottom line; it varies.
For answering this question we have to consider the constant voltage drop model of the diode which says that if voltage across diode is less then its cut in voltage than assume diode to be open circuit and if it is greater then assume diode to be short circuit.Till the input voltage is less than the cut in voltage, diode is open circuit(thus no current through the circuit). Thus entire input voltage appears across the diode as output.When input voltage is greater than or equal to cut in voltage, then short circuit the diode. Thus, there will be no voltage drop across the diode as output.Thus cut in voltage decides when to consider the diode open circuit and when short circuit. It decides when the diode will have output when it will not.
The effect of diode voltage drop as the output voltage is that the input voltage will not be totally transferred to the output because power loss in the diode . The output voltage will then be given by: vout=(vin)-(the diode voltage drop).
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).
zener diode
zener diode :zener diode operates under reverse bias voltageideal diode :ideal diode operates under forward bias voltage
forward drop is the same as any other silicon diode, about 0.7V
voltage drop deviding accure
A silicon diode has a voltage drop of approximately 0.7V, while a germanium diode has a voltage drop of approximately 0.3V. Though germanium diodes are better in the area of forward voltage drop, silicon diodes are cheaper to produce and have higher breakdown voltages and current capabilities.
Yes, the forward voltage drop of a Schottky diode is usually more than the forward voltage drop of a tunnel diode. A Schottky diode voltage drop is between approximately 0.15 to 0.45 volt. The interesting thing that makes a tunnel diode different from other diodes is its "negative resistance region" with a "peak current" around 0.06 volt and a "valley current" around 0.30 volt.
Consider ideal diode to be connected in series with resistor of 6kSilicon diode forward bias voltage = 0.7 voltsCurrent across 6k resistor = (5-0.7)/6000 amperesVoltage across {resistor + diode}=4.3 + 0.7=5vIf silicon internal resistance is 6k then voltage across diode=5vIf external resistance is 6k and diode resistance is negligible then voltage across diode=0.7v
The forward biased voltage drop of a diode depends on the type of diode and the current through the diode. A typical silicon diode will exhibit a voltage drop between 0.6v and 1.4v depending on current. An LED might range from 2v to 3v. A germanium diode might go a low as 0.2v. Bottom line; it varies.
For answering this question we have to consider the constant voltage drop model of the diode which says that if voltage across diode is less then its cut in voltage than assume diode to be open circuit and if it is greater then assume diode to be short circuit.Till the input voltage is less than the cut in voltage, diode is open circuit(thus no current through the circuit). Thus entire input voltage appears across the diode as output.When input voltage is greater than or equal to cut in voltage, then short circuit the diode. Thus, there will be no voltage drop across the diode as output.Thus cut in voltage decides when to consider the diode open circuit and when short circuit. It decides when the diode will have output when it will not.
An ideal diode:Passes current in one direction only. (Under forward bias).Has no leakage current (passes no current under reverse bias).Has no forward voltage drop. (No voltage loss under forward bias - a real diode has Vd~=0.7)See links for more details.