That's to avoid the diode from burning out. When it is forward biased, the resistance is extremely low; so even a small voltage can produce a large current, and that can quickly damage it.
To forward bias a germanium diode you need to reach between 0.2 and 0.3 V between anode (+) and cathode (-). Once this voltage is reached, the diode will conduct. Make sure you have a resistor to limit the forward current to a safe (for the diode, that is) value.
An example of a resistor that transforms electrical energy to light energy is a light-emitting diode (LED). When an electric current passes through an LED, it emits light as a result of the energy conversion process within the semiconductor material of the LED.
The unit of a diode is the ampere (A), which measures the flow of electric current through the diode. It indicates the amount of current passing through the diode at a given moment.
A PN Junction Diode is one of the least complex Semiconductor Devices around, and which has the normal for passing current in one and only heading just. Be that as it may, dissimilar to a resistor, a diode does not carry on directly regarding the connected voltage as the diode has an exponential current-voltage ( I-V ) relationship and thusly we can not portrayed its operation by just utilizing a comparison. ex: Ohm's law.
-- The definition of 'reverse bias' is anode negative with respect to the cathode, or negative voltage across the diode. That places the graph in negative-x territory. -- The diode simply acts as a resistor. Its unique 'diode' characteristics arise from the fact that its 'resistance' changes with different bias points, but the current through it always has the same polarity as the voltage across it. Therefor . . . -- When the voltage across it is negative, the current through it is also negative. Negative current appears on the graph in negative-y territory. -- Negative-x territory/negative-y territory is the third quadrant.
There isn't anything "scientific" about this simple series circuit. If you've got a variable resistor (a potentiometer) and a diode in series, you have a simple series circuit with the two components in it.
The diode voltage drop is 0.7 volts, so you need that much to turn it on. Current is controlled by a resistor in series.
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
No... Diode block one way... resistor block eitherway.
An inductor will supply better current source.
If diode is connected in series then current will flow only in one direction. ie. Current flow occurs only when diode is forward biased. ANSWER: It depends are the diodes are connected in series if they are back to back no current will flow if connected in the forward conduction mode then they will conduct.
The junction (diode or transistor) will be destroyed.
esistors restrict the flow of electric current, for example a resistor is placed in series with a light-emitting diode (LED) to limit the current passing through the LED.
Because Thevenin does not work for a nonlinear network, e.g. a battery in series with a diode and resistor.
None. A diode does not have a resistor in it. Diodes and resistors are different devices used for different purposes. For example, increasing current flow through a resistor increases the voltage drop across the resistor. Increasing current flow through a diode (within the diode's capabilities, just the same as a resistor) will actually decrease the voltage drop across the diode (once the diode "turns on" this voltage variance will be fairly small, though).
you could use a current limiting diode such as a Zenner, or a current limiting resistor in series with the LED
Transistor Diode Resistor Capacitor