On Monday 22 December 2008 04:07:10 am Mike Hatch wrote:
A simple
example of the differnce. If you take a D-type flip-flop and
connect the output to a NOT gate and the output of that back to the D
input, then you get a divide-by-2-circuit. Feed in a regular clock at one
frequency and the Q output will toggle at half that frequency. But if you
A clocked JK with the JK held high will perform the same function.
As will tying the D input back to not-Q, which most of them have. The
advantage to using the D-type over the J-K is two fewer package pins needed,
which may or may not be important to somebody.
And what the heck does that J-K stand for, anyway?
----- Original Message -----
From: "Tony Duell" <ard at p850ug1.demon.co.uk>
To: <cctalk at classiccmp.org>
Sent: Saturday, December 20, 2008 6:10 PM
Subject: [personal] Re: PDP-8 on FPGA project & where is Hans Pufal?
>> > Your code above does not result in latches. It results in D type
>> > flipflops!
>>
>> What's the difference? As I've learned to use the words, D-flops _are_
>> latches. Have I mislearnt?
>
> As I understand the terms, a D-type flip-flop is edge-triggered, a latch
> (or more exactly a transparent latch) is level-operated.
>
> What I eman is that for a D-type flip-flop, the output (Q) is set equal
> to the input (D) a short time after the rising edge (say), of the clock
> signal. At all othter times Q does not change state, no matter what D
> does. So if clock is held high all the time, or held low all the time, Q
> will never change.
>
> But with a transparent latch, then if the clock (sometimes called the
> enable input) is in one sate (say high). then the output (Q) tracks the
> input (D). When the clock input goes low, then Q is held in whatever
> state it was in as the clock input went low.
>
> A simple example of the differnce. If you take a D-type flip-flop and
> connect the output to a NOT gate and the output of that back to the D
> input, then you get a divide-by-2-circuit. Feed in a regular clock at one
> frequency and the Q output will toggle at half that frequency. But if you
> do the same thing with a transparent latch, you get something that
> oscillates at a frequency determined by the propagation delays when the
> clock is high. In other words the output is somewhat unpredicatable, it's
> certainly not a square wave at half the frequency of the clock signal.
>
> -tony
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