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What else can I help you with?
Why is a resistance usually connected in series with zener diode?
To drop the unwanted voltage.
How much forward diode voltage is there with the ideal diode approximation?
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
What happens in the circuit when a1N4001 diode experiences its breakdown voltage of 50V?
If a diode breaks down in the reverse direction then, providing there is current-limiting resistance, nothing will happen to the diode. If there is no such current-limiting resistance the diode will be destroyed. Heat is the only enemy of a semiconductor of this type.
What is the current of a circuit if the voltage is 10V with a 1.2K Resistor and a diode?
you should specify: - circuit topology, I assume a series connection. - diode allows current flow? It depends how it's connected - diode forward voltage drop value if diode is in forward conduction, you have VR=10V - VDIODE and, thus, I = VR/R=(10-Vdiode)/1200.
What is the effect of diode voltage drop as the output voltage?
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).
Is it necessary to add a series resistance with the diode or not?
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
Why is a resistance usually connected in series with zener diode?
To drop the unwanted voltage.
How much forward diode voltage is there with the ideal diode approximation?
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
How does the cut in voltage effects the output of a diode clipper?
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.
How does the ac resistance of a diode change as the voltage increases?
When the voltage increases the temperature in the diode also increases. When the temperature in the diode increases, the resistance decreases.
How do you change the amount of voltage in a circuit?
Any transformer can do it Even a coil or diode cap pumps. or increasing the resistance
Bulk resistance in diode?
the amount of voltage applied to the diode
What resistance does a practical diode offer under reverse-biased conditions if the applied reverse voltage is less than the breakdown voltage?
A practical diode under reverse-biased conditions, if the applied reverse voltage is less than the breakdown voltage, will offer a high resistance, usually in excess of 10 megohms. In a practical circuit, the diode would appear to be open.
What happens in the circuit when a1N4001 diode experiences its breakdown voltage of 50V?
If a diode breaks down in the reverse direction then, providing there is current-limiting resistance, nothing will happen to the diode. If there is no such current-limiting resistance the diode will be destroyed. Heat is the only enemy of a semiconductor of this type.
Diode reverse breakdown14v current of 70mA and an un-stabilised supply voltage of 28v calculate resistance?
If a circuit element has a voltage of 14V and a current of 70mA, then the resistance of the circuit element is 200 ohms. This is ohm's law. The resistance or type of the power supply is meaningless.
Infinite resistance in one direction of Diode?
After isolating the diode from the circuit, the diode shows infinite resistance in one direction and low resistance in the other direction. The diode
What is the typical value of the barrier potential for a silicon diode?
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.
