Timeline for Understanding the rigorous definition of hold time
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S Jan 15 at 21:19 | history | bounty ended | EE18 | ||
| S Jan 15 at 21:19 | history | notice removed | EE18 | ||
| Jan 12 at 16:22 | history | edited | EE18 | CC BY-SA 4.0 |
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| Jan 12 at 16:21 | vote | accept | EE18 | ||
| Jan 11 at 23:56 | answer | added | W5VO | timeline score: 3 | |
| S Jan 11 at 20:49 | history | bounty started | EE18 | ||
| S Jan 11 at 20:49 | history | notice added | EE18 | Draw attention | |
| Jan 9 at 16:56 | comment | added | EE18 | @Andyaka As above, I'd also appreciate hearing what you think about this edit. | |
| Jan 9 at 16:56 | comment | added | EE18 | @jsotola I have written a lengthy edit and would appreciate hearing what you think. Thank you! | |
| Jan 9 at 16:55 | history | edited | EE18 | CC BY-SA 4.0 |
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| Jan 9 at 16:23 | comment | added | EE18 | I agree heuristically, but as you can see they give a definition which is a little bit more elaborate than that. It seems to be "the length of time D has to remain stable after the clock such that, even if the flip flop captures the token, the \$t_{CQ}\$ is greater than \$t_{pcq}\$". @Andyaka | |
| Jan 9 at 16:19 | comment | added | Andy aka | @EE18 I believe it means the length of time D has to remain stable after clock has occurred. | |
| Jan 9 at 16:17 | comment | added | EE18 | @jsotola I agree, but (calling this \$t_{CD}\$ since they don't give notation to it) wouldn't we arrive at the conclusion I wrote to Andy above? I must be missing something. | |
| Jan 9 at 16:16 | comment | added | EE18 | Are you saying that \$t_{hold}\$ is defined as the minimum \$t_{DC}\$ (this is D to clock per the above) which is such that the corresponding \$t_{CQ}\$ (which is a function of \$t_{DC}\$ is less than or equal to \$t_{pcq}\$. Per the Figure 10.35, wouldn't this definition lead to \$t_{hold} = -t_{setup}\$ (the minus sign coming from the definition of the positive direction for hold time) since for any \$t_{DC} < t_{setup}\$ we have \$t_{CQ} > t_{pcq}\$ (basically the \$t_{CQ}(t_{DQ})\$ function is monotonic)? @Andyaka | |
| Jan 9 at 16:14 | comment | added | jsotola |
it's not clock to D ... it is clock to changing of D
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| Jan 9 at 16:08 | comment | added | Andy aka | I expect "clock to D" is the hold time. They seem to have chosen an awkward way of stating things generally. | |
| Jan 9 at 16:01 | history | asked | EE18 | CC BY-SA 4.0 |