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As I recall, it used to be in the U.S. that the ability to read and send Morse code over CW was a requirement for an amateur license (and was tested, I believe). Is that still true in the U.S.? If not when was it dropped? Around the world, is Morse a requirement for amateur licensing?

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    $\begingroup$ Around the world is a little broad (+: You might consider narrowing it down to particular nations $\endgroup$ – VU2NHW Oct 30 '13 at 20:13
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    $\begingroup$ Separate answers per country seems useful to me. Agree? $\endgroup$ – Warren VE3WPX Jan 27 '14 at 16:32
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    $\begingroup$ @WarrenP if each answer is a separate country, by what basis do you upvote and accept them? Separate countries per question seems more useful to me. This way, the questions can be tagged by their country, so if I'm just searching for licensing in Canada, I can search "[license] [canada]". $\endgroup$ – Phil Frost - W8II Feb 5 '14 at 12:57
  • $\begingroup$ If the Beta Meta Gods so declare, then I'll go with that. More I think about it, more I like your idea. $\endgroup$ – Warren VE3WPX Feb 6 '14 at 21:11
  • $\begingroup$ relevant meta discussion: meta.ham.stackexchange.com/questions/120/… $\endgroup$ – Warren VE3WPX Feb 6 '14 at 21:16
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It is not required any more for any US license, as of 2008. It has not been required for the technician license since 1991. It is still required in some areas, and some international amateur radio licenses, in particular the IARP license, which may be used by US amateurs to operate outside of the US in IARP countries.

Bottom line is, however, that fewer and fewer countries are requiring it, ever since the 2003 ruling from the ITU that no longer required knowledge of CW for operating on HF.

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In Australia, the requirement for Morse was dropped when the new licence structure of three classes (Foundation, Standard and Advanced) was adopted. This happened in October 2005.

Since that date there has been no requirement for any Australian amateur to demonstrate proficiency in Morse for any class of licence.

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Apart from Licensing, CW is required for using some specific frequencies. For a U.S. list of such frequencies see this ARRL page: http://www.arrl.org/band-plan

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  • $\begingroup$ band plans have no force of law, do they? This is "the code of the tribe" but not quite the law of the land. $\endgroup$ – Warren VE3WPX Feb 4 '14 at 2:28
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Here in VU land i.e. India following the 2005/2009 notifications from WPC (Wireless Planning & Coordination Wing), CW is not required for the Indian equivalent of the combined US novice/tech called 'restricted'.

A nominal 8WPM CW is however required for the Indian equivalent of US extra/general.

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In Canada, CW Morse code skills is not required, but you may elect to seek a CW exam, which confers the same honors now as the "With Honors Basic" qualification, to grant the same priveleges as you would obtain if you simply passed the Honors qualification with 80% or more. Practically speaking there is almost zero reason to try to pass the CW exam in Canada, since it's easy to get over 80% on the Basic certificate.

Amateur radio operator certificates and their rules are overseen by Industry Canada, formerly DOC (department of Communications). Rules and priveleges and CW requirements are discussed here.

Their site is hard to navigate but the gist of their current rules is that a Basic certificate has no mandatory CW requirements, but passing a CW (Morse code) test formerly conferred additional privileges. As of 2005, Basic certificates confer all the privileges, IF you pass with 80% or more.

Sources: IC, and this book: Canadian Amateur Radio Basic Qualification Study Guide, 8th edition, Coax Publications Inc. (http://www.coaxpublications.ca/)

Update: Formerly I had said you needed it for Advanced qualifications. You do not.

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  • $\begingroup$ Can you list sources? $\endgroup$ – Seth Jan 27 '14 at 17:03
  • $\begingroup$ No, there is no requirement in Canada to have CW for the Advanced certificate. Please see RIC-3 §1.4 “Privileges and Restrictions $\endgroup$ – scruss Feb 1 '14 at 18:35
  • $\begingroup$ Sorry I edited my answer. better now? $\endgroup$ – Warren VE3WPX Feb 4 '14 at 2:28

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