In the cut-off region of a transistor, the base-emitter junction is not forward-biased, meaning the transistor is effectively turned off. As a result, no current flows from collector to emitter, and the voltage across the collector to emitter (V_CE) is approximately equal to the supply voltage (V_CC) connected to the collector. Therefore, V_CE is at its maximum value, close to V_CC, indicating that the transistor is not conducting.
If the signal frequency exceeds the cut-off frequency in a waveguide, the wave will not propagate through the waveguide and will be attenuated. This is because the waveguide is designed to support only certain modes of propagation, and frequencies above the cut-off do not satisfy the waveguide's mode conditions. Instead, the signal will either reflect back or be absorbed, leading to a loss of signal strength.
yes , it has high resistance between collector and emitter on the off state.
Cut off frequency is that frequency at which the power output of the circuit reduces to half of its power in pass band. Power (P)= Current2 X Resistance. so to reduce the power by half we need to reduce the current by 1/sqrt{2} = 0.707 = 3 db. So it's also called half power frequency. http://enter4joy.com
All amplifier typically exhibit a band-pass frequency response. The cut off frequency in the low end is usually determined by the coupling band bypass capacitor .and the high frequency limit is typically determined by internal capacitances in the transistor itself.
Collector-to-Emitter resistance is high when the transistor is biased off.
In the cut-off region of a transistor, the base-emitter junction is not forward-biased, meaning the transistor is effectively turned off. As a result, no current flows from collector to emitter, and the voltage across the collector to emitter (V_CE) is approximately equal to the supply voltage (V_CC) connected to the collector. Therefore, V_CE is at its maximum value, close to V_CC, indicating that the transistor is not conducting.
In response to the control signal on the base, the transistor switches on or off, effectively shorting the collector to the grounded emitter or opening leaving the collector floating.
A transistor can be in three conditions or states. It can be active (at a voltage higher than the emitter), in saturation or cut off (no current).
If the signal frequency exceeds the cut-off frequency in a waveguide, the wave will not propagate through the waveguide and will be attenuated. This is because the waveguide is designed to support only certain modes of propagation, and frequencies above the cut-off do not satisfy the waveguide's mode conditions. Instead, the signal will either reflect back or be absorbed, leading to a loss of signal strength.
Because it has the lowest cut-off frequency (highest cut off wavelength) for a>b o
Because of the geometry of the common collector configuration, changes in base voltage appear at the emitter. Said another way, what happens at the base pretty much happens at the emitter, and the emitter can be said to "mirror" or "follow" the base. The emitter is a follower of the base, and the name emitter follower appeared and was used.
yes , it has high resistance between collector and emitter on the off state.
Cut off frequency is that frequency at which the power output of the circuit reduces to half of its power in pass band. Power (P)= Current2 X Resistance. so to reduce the power by half we need to reduce the current by 1/sqrt{2} = 0.707 = 3 db. So it's also called half power frequency. http://enter4joy.com
The beta cut off frequency is the frequency at which the current amplification of an amplifier transistor drops to three decibels below its value at 1 kilohertz. This is used in electronic engineering.
The common answer is that Perseus cut off her head.
yes it can have coz fc= c/2[(m/a)2+(n/b)2]1/2 therefore for various modes(which decide m,n) nd dimensions(a,b) of waveguide it will have different cut off frequency. A normal waveguide is used in the octave frequency range where only the fundamental mode can propagate.