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Define pinch off voltage and the gate source cut off voltage?
If negative voltage is applied to the gate of a NMOS, it repels electrons from the channel region towards the bulk of the p-substrate and attaract holes from p-substrate towards the channel. The recombination between holes and electrons causes a deplation of majority carriers in the channel. Enough nagative gate voltage can cause the channel depleted of majority carriers and cuts off the current between the source and the drain. The least negative gate voltage causing this is called gate-source cut off voltage.
What are depletion and enhancement mosfet?
A depletion MOSFET is a MOSFET that is normally on. It outputs maximum current when the gate-source voltage is 0V. As the gate-source voltage increases, the drain-source channel becomes more resistive and the current decreases. An enhancement MOSFET has the opposite behavior. It is normally off. It outputs no current when the gate-source voltage is 0V. As the gate-source voltage increases, the drain-source channel becomes less resistive and the current increases.
Why current flow from drain to source in JFET?
In a Junction Field-Effect Transistor (JFET), current flows from the drain to the source due to the application of a voltage between these terminals, creating an electric field that allows charge carriers (electrons for n-channel JFETs or holes for p-channel JFETs) to move through the channel. The gate voltage controls the channel's conductivity by modulating the width of the depletion region, which affects the flow of charge carriers. When the drain-source voltage (V_DS) is applied, it causes electrons to flow from the drain to the source in an n-channel JFET, completing the circuit. The flow direction is thus determined by the polarity of the applied voltages and the type of semiconductor material used.
If the resistance increases what will happen to voltage and current?
If you have a simple circuit. For eg: One voltage source and one resistor, then the voltage of the circuit will always remain the same, the current however will decrease following Ohms' Law V=I*R. If we have a current source instead of a voltage source, we are forcing the current to be a certain value so if we increase the resistor value the current will remain the same but the voltage will increase.
2 What is Pinch-off voltage of JFET?
To find out the qualitative response to changes in Vgs and Vds, consider Vgs =0. In response to a small applied voltage Vds, a n-channel JFET acts as a simple semiconductor resistor and Id increases linearly with Vds. With increasing current, the ohmic voltage drop between the source and the channel region reverse biases the junction and the conducting portion of the channel begins to constrict. because of the ohmic drop along the length of the channel itself, the constriction is not uniform but is more pronounced at distances father from the source. Eventually, a Vds is reached at which the channel is "pinched off". At this voltage, Id begins to level off. and approach a constant value. It is not possible for the channel to close completely and reduce Id to zero. If that were the case, then the ohmic drop required to provide the back biasing would be lacking.Id is not equal to zero at Pinch off voltage.Thus, pinch off voltage can be defined as the gate reverse voltage that removes all the free charge from the channel.
Define pinch off voltage and the gate source cut off voltage?
If negative voltage is applied to the gate of a NMOS, it repels electrons from the channel region towards the bulk of the p-substrate and attaract holes from p-substrate towards the channel. The recombination between holes and electrons causes a deplation of majority carriers in the channel. Enough nagative gate voltage can cause the channel depleted of majority carriers and cuts off the current between the source and the drain. The least negative gate voltage causing this is called gate-source cut off voltage.
What is always true during the operation of an electrolytic cell?
the negative terminal of the voltage source is attached to cathrode.
Why source is more negative than gate in JFET?
In a Junction Field Effect Transistor (JFET), the source is typically more negative than the gate because the source terminal is connected to the channel, which allows current to flow through it. The gate, being reverse-biased, creates an electric field that controls the channel conductivity. This reverse bias means that the gate voltage is generally lower (more negative) than the source voltage, which allows for proper operation of the JFET by maintaining a depletion region that modulates current flow. Thus, the negative potential at the source helps establish the necessary conditions for the JFET to function effectively.
What is the threshold value of the PMOS?
The threshold voltage (V_th) of a PMOS transistor is the gate-source voltage (V_GS) at which the transistor begins to conduct. It is typically negative, meaning the gate voltage must be lower than the source voltage to create a conductive channel. The exact value of the threshold voltage can vary based on the specific technology and fabrication process, but it generally ranges from -0.5V to -2V for most PMOS devices.
What does the voltage used by each part of a circuit always adds up to?
A: the source voltage
What is Cut-off voltage in fet?
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 in cut off voltage?
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
What are depletion and enhancement mosfet?
A depletion MOSFET is a MOSFET that is normally on. It outputs maximum current when the gate-source voltage is 0V. As the gate-source voltage increases, the drain-source channel becomes more resistive and the current decreases. An enhancement MOSFET has the opposite behavior. It is normally off. It outputs no current when the gate-source voltage is 0V. As the gate-source voltage increases, the drain-source channel becomes less resistive and the current increases.
Why do you always use voltage source in lab not current source?
Current is created by voltage (potential difference), not the other way round. It's the voltage - or 'push' - that is making your current move.
Why peak voltage of the load is smaller than the voltage from the transformer output?
The voltage appearing across a load is always smaller than the no-load voltage of any voltage source -e.g. batteries, generators, or transformers. In simple terms this is because all these voltage sources have internal resistance or impedance which results in an internal voltage drop when the source delivers a load current. The resulting voltage, therefore, is always the difference between the no-load voltage and the internal voltage drop. A measure of the difference between a source's no-load and full-load voltage is termed its 'voltage regulation'.
What will happen when the terminal of the volmeter are inversed and used to read the voltage of the DC supply?
Depends on the voltmeter. Some meters will display a positive voltage no matter how the leads are attached to the source. Others may indicate a negative voltage if leads are reversed.Another AnswerAn analogue voltmeter will always read downscale (i.e. 'backwards') if connected the wrong way around.Digital voltmeters generally indicate the correct voltage, but display a negative sign to indicate that the polarity is the wrong way around.Don't forget, when we say 'negative voltage', we are referring to its direction. Voltage, which is simply another word for 'potential difference', and it cannot be positive or negative in the sense of 'charge'.
Why current flow from drain to source in JFET?
In a Junction Field-Effect Transistor (JFET), current flows from the drain to the source due to the application of a voltage between these terminals, creating an electric field that allows charge carriers (electrons for n-channel JFETs or holes for p-channel JFETs) to move through the channel. The gate voltage controls the channel's conductivity by modulating the width of the depletion region, which affects the flow of charge carriers. When the drain-source voltage (V_DS) is applied, it causes electrons to flow from the drain to the source in an n-channel JFET, completing the circuit. The flow direction is thus determined by the polarity of the applied voltages and the type of semiconductor material used.
