- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
I would like to ask a quick question regarding the DFFE connection in Quartus.
As I observed in the symbol, there is a bubble (inverter) before both the PRN and CTRN inputs.
In this case, if I connect a High signal to PRN or CTRN, will the internal PRN and CTRN signals stay High, or will they become Low due to the inverter?
If there are any official documents, application notes, or recommended practices you could share on this topic, I would be grateful.
Thank you very much for your support.
Link Copied
3 Replies
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Hi,
They will become low due to inverter because they're active low.
Thanks,
Regards,
Sheng
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
Hi,
May I know do you have any further concern or consideration?
Thanks,
Regards,
Sheng
- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Permalink
- Report Inappropriate Content
"As I observed in the symbol, there is a bubble (inverter) before both the PRN and CTRN inputs.
In this case, if I connect a High signal to PRN or CTRN, will the internal PRN and CTRN signals stay High, or will they become Low due to the inverter?"
No. The circle doesn't indicate an inverter before "internal PRN". It indicates active low function of PRN, also letter N does the same. But it's not an inverter before an active low input.
Connecting high respectively VCC to an active low input deactivates the input, leaving the pin open does the same.
In this case, if I connect a High signal to PRN or CTRN, will the internal PRN and CTRN signals stay High, or will they become Low due to the inverter?"
No. The circle doesn't indicate an inverter before "internal PRN". It indicates active low function of PRN, also letter N does the same. But it's not an inverter before an active low input.
Connecting high respectively VCC to an active low input deactivates the input, leaving the pin open does the same.

Reply
Topic Options
- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page