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!
One of the simplest circuits is the asynchronous or ' ripple' counter. Below is shown the circuit diagram of a simple 3 stage ripple counter.
The operation of this circuit is based on the fact that the truth table for the JK flip flop is only valid if the clock waveform is falling, i.e. 1->0. Assume the outputs are all zero, the flip flops will not change until the clock on each flip flop falls. The clock in waveform has just fallen ,since the JKa inputs are logic '1' the device will toggle and the output will invert i.e. Qa=1. Flip flop B will not change because the clock waveform on B has risen (0->1) and these devices only functions on a falling edge. The clock in waveform has fallen again, so Qa toggles again (i.e. Qa =0), this has just produced a falling clock on JKb and Qb toggles (i.e. Qab=1) .The device has just counted from 000-> 001->010.
The circuit is called a ripple counter because the clock pulse is slowly rippling through the JK's, hence asynchronous (Not at the same time!) .The limitations of the asynchronous counter is the speed of operation. A rough formula for the maximum speed is when the clock changes before the output changes i.e. F = 1 / n x propagation delay where n = number of stages, propagation delay of one JK
A better technique is to use a synchronous design where all the JK are clocked together so the maximum frequency is only limited by the propagation delay of 1 JK.
The circuit appears to be complex in design, however it is easily realised by using state diagrams. The maximum frequency of operation is again roughly calculated by considering the frequency at which the output just changes before the clock in changes. F = 1/ Propagation delay
determine the efficience of a 16 KVA transformer for the full-load,unit power factor
Q. Point charges, each of √4πε 0 C, are located at the vertices of an equilateral triangle of side a. Determine the electric force on each charge.
Registers Various registers shown in figure are discussed below in detail.
Give the examples of sinusoidal with two equations?
Q. A six-pole, double-layer dc armature winding in 28 slots has five turns per coil. If the field flux is 0.025 Wb per pole and the speed of the rotor is 1200 r/min, find the value
Using Bode plot calculate (a) Phase margin (b) Gain margin (c) Stability of closed loop system. The open transfer function of the system is t=30/(s+2)/(s+3)Using Bode plot calcula
maxium capacity of genrators in practical use
In this case, 24 simultaneous calls can be put through switch. Generally a 24-outlet Uni-selector is used as a selector hunter. Every one of the 24 outlets is connected to one two-
Q. A four-pole dc machine with 728 active conductors and 30mWb flux per pole runs at 1800 r/min. (a) If the armature winding is lap wound, find the voltage induced in the armatu
can voltage regulation be negative?why?
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: +1-415-670-9521
Phone: +1-415-670-9521
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd