Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
Q. What do you mean by Transconductance?
The control that the gate voltage has over the drain current is measured by transconductance and is similar to the transconductance of the tube.It may be defined as follows:
It is is the ratio of the change in drain current (ΔId) to the change in gate source voltage(Vgs) at constant drain source voltage that is
Transconductance = Id / Vgs at constant Vds.
The transconductance of a JFET is usually expressed either in mA or microamperes.
Amplification factor(µ)
It is the ratio of change in drain source voltage to the change in gate source voltage at constant drain current
Amplification factor= ΔVds/ ΔVgs at constant Id
Amplification factor of a JFET indicates how much more control the gate voltage has over drain current than has the drain voltage.
Relation among JFET parameters.
The relationship among JFET parametes can be established as under:
We know µ=Vds/ Vgs
Hence
µ = rd*gfs.
Amplification factor = a.c drain resistance * transconductance.
A single phase line has an impedance of 8.4 + j11.2 ?. The line feeds a load consisting of a resistor and an inductor connected in parallel as shown in Figure 1. The load is absorb
Q. Explain function of application layer? Layers of OSI model are as follows: (1) The Physical Layer: This defines an interface in terms of connections, voltage levels and
Serial Input output Ports For serial input and output there are two pins in 8085 microprocessor
Q. A transmitter is connected to an antenna by a transmission line for which ¯ Z 0 = R 0 = 5 0 . The transmitter source impedance is matched to the line, but the antenna is kno
Find the current flow through resistor 12 Ω using Thevenin's Theorem.
Q. A20-hp, 250-Vshuntmotor has a total armature - circuit resistance of 0.25 and a field-circuit resistance of 200.At no load and rated voltage, the speed is 1200 r/min, and the
The Parallel Resistance rule Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4
Determine the total energy loss: Two capacitors C 1 = 50 μF and C 2 = 100 μF are connected in parallel across 250 V supply. Determine the total energy loss. Figure
1) Consider the magnetic circuit shown in the figure. Steady currents flow in the windings of N 1 and N 2 turns on the outside legs of the ferromagnetic core. The core has a
Disadvantages
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd