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
Can we use ohm''s law for AC CIRCUITS?
1. Introduction : Theory: The voltage measured across a load follows the Ohm's law which says that the current passing through a conductor between two points is direct
Q. A balanced wye-connected load with a per-phase impedance of 4 + j3 is supplied by a 173-V, 60-Hz three-phase source. (a) Find the line current, the power factor, the total
Refer to Figure 100. Assume MKS units. Given: R1= 4, R2=14, R3= 9, I4= 8, I5= 7. Determine: Ieq, Req, and V3.
Analysis Flow Similar to most subjects, the analysis of semiconductor devices is also performed by starting from simpler problems and regularly progressing to more complex one
Multiprogramming - Single Processor one Users Many Programs When single processor is used to execute more than one independent program simultaneously the technique is ca
Network Reconductoring - Technical Loss Reduction The size of conductor/cable is a significant parameter as it determines the current density and the resistance of the line. A
a short review article on the classification of time frame
LDAX Load Accumulator Indirect Instruction This instruction is used to copy data from memory location pointed by register pair only BC or DE to the accumulator HL pair
Q. What do you mean by Lumped-circuit elements ? Electric circuits or networks are formed by interconnecting various devices, sources, and components. Although the effects of e
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