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	<title>Comments on: 06. Oscillator and Timing</title>
	<atom:link href="http://www.engscope.com/pic24-tutorial/6-oscillator-and-timing/feed/" rel="self" type="application/rss+xml" />
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	<description>Engineering, Kayaking, Guitar and More</description>
	<lastBuildDate>Wed, 01 Sep 2010 00:29:56 +0000</lastBuildDate>
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		<title>By: Danny Cranmer</title>
		<link>http://www.engscope.com/pic24-tutorial/6-oscillator-and-timing/comment-page-1/#comment-7146</link>
		<dc:creator>Danny Cranmer</dc:creator>
		<pubDate>Sat, 06 Mar 2010 20:06:04 +0000</pubDate>
		<guid isPermaLink="false">http://www.engscope.com/oscillator-and-timing/#comment-7146</guid>
		<description>After Testing my code on a development board I made and found CN0 CN1 working I realised that I was being stupid and hadn&#039;t made a contact on them 2 pins whilst soldering,,, should have been the first thing I checked! 

About the SPI errata stuff I had a quick look,, there is a lot of stuff about the SPI but could not see the problem I encountered, but my work around is working fine so ill just stick with that for now.

Thanks

Danny</description>
		<content:encoded><![CDATA[<p>After Testing my code on a development board I made and found CN0 CN1 working I realised that I was being stupid and hadn&#8217;t made a contact on them 2 pins whilst soldering,,, should have been the first thing I checked! </p>
<p>About the SPI errata stuff I had a quick look,, there is a lot of stuff about the SPI but could not see the problem I encountered, but my work around is working fine so ill just stick with that for now.</p>
<p>Thanks</p>
<p>Danny</p>
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	<item>
		<title>By: jliu83</title>
		<link>http://www.engscope.com/pic24-tutorial/6-oscillator-and-timing/comment-page-1/#comment-7051</link>
		<dc:creator>jliu83</dc:creator>
		<pubDate>Tue, 23 Feb 2010 00:06:16 +0000</pubDate>
		<guid isPermaLink="false">http://www.engscope.com/oscillator-and-timing/#comment-7051</guid>
		<description>Make sure you have the secondary oscillator disabled, this is usually found in the configuration bits.  Also, scan through the datasheet for the SOSC0 pin, there might be more clues as to why you are getting problems.  My guess is that the secondary oscillator module is enabled, and it is actually waiting for a secondary clock on the pins.

Good luck, tell me how it goes.

-J</description>
		<content:encoded><![CDATA[<p>Make sure you have the secondary oscillator disabled, this is usually found in the configuration bits.  Also, scan through the datasheet for the SOSC0 pin, there might be more clues as to why you are getting problems.  My guess is that the secondary oscillator module is enabled, and it is actually waiting for a secondary clock on the pins.</p>
<p>Good luck, tell me how it goes.</p>
<p>-J</p>
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	<item>
		<title>By: Danny Cranmer</title>
		<link>http://www.engscope.com/pic24-tutorial/6-oscillator-and-timing/comment-page-1/#comment-7043</link>
		<dc:creator>Danny Cranmer</dc:creator>
		<pubDate>Sun, 21 Feb 2010 23:30:05 +0000</pubDate>
		<guid isPermaLink="false">http://www.engscope.com/oscillator-and-timing/#comment-7043</guid>
		<description>Hi I&#039;m using a PIC24FJ128GA006 for a student project, basically I am using the Interrupt on change function and cannot get RC13 / RC14 ( CN0 / CN1 ) to work which happen to be SOSC0/SOSCI, I have managed to use interrupt on change pins on PORTG. Have set TRIS reg and am using internal fast RC, if required i can post code but ive been through datasheet many times and cannot find much about this. 

Have tried just using as a general IO controlling an LED too but no go.

thanks

Danny</description>
		<content:encoded><![CDATA[<p>Hi I&#8217;m using a PIC24FJ128GA006 for a student project, basically I am using the Interrupt on change function and cannot get RC13 / RC14 ( CN0 / CN1 ) to work which happen to be SOSC0/SOSCI, I have managed to use interrupt on change pins on PORTG. Have set TRIS reg and am using internal fast RC, if required i can post code but ive been through datasheet many times and cannot find much about this. </p>
<p>Have tried just using as a general IO controlling an LED too but no go.</p>
<p>thanks</p>
<p>Danny</p>
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	<item>
		<title>By: jliu83</title>
		<link>http://www.engscope.com/pic24-tutorial/6-oscillator-and-timing/comment-page-1/#comment-1859</link>
		<dc:creator>jliu83</dc:creator>
		<pubDate>Wed, 27 May 2009 20:32:26 +0000</pubDate>
		<guid isPermaLink="false">http://www.engscope.com/oscillator-and-timing/#comment-1859</guid>
		<description>Hi Carl,
  You forgot that Fcy is always calculated with a 1/2 factor with respect to Fin.  Each instruction cycle requires two oscillator cycles.  So the correct value would be 8 x 10 / (2 * 2 * 2) = 10 Mhz (assuming DOZE =  FYC/1), with PLLFBD = 0x08.  There was a mistake in the code that I did not catch.  I got the &quot;+2&quot; as a &quot;-2&quot;.  PLLFBD should be 0x06, with M = 8, as has been corrected right now. 

Thanks,
-J</description>
		<content:encoded><![CDATA[<p>Hi Carl,<br />
  You forgot that Fcy is always calculated with a 1/2 factor with respect to Fin.  Each instruction cycle requires two oscillator cycles.  So the correct value would be 8 x 10 / (2 * 2 * 2) = 10 Mhz (assuming DOZE =  FYC/1), with PLLFBD = 0&#215;08.  There was a mistake in the code that I did not catch.  I got the &#8220;+2&#8243; as a &#8220;-2&#8243;.  PLLFBD should be 0&#215;06, with M = 8, as has been corrected right now. </p>
<p>Thanks,<br />
-J</p>
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	<item>
		<title>By: Carl</title>
		<link>http://www.engscope.com/pic24-tutorial/6-oscillator-and-timing/comment-page-1/#comment-1854</link>
		<dc:creator>Carl</dc:creator>
		<pubDate>Wed, 27 May 2009 17:05:33 +0000</pubDate>
		<guid isPermaLink="false">http://www.engscope.com/oscillator-and-timing/#comment-1854</guid>
		<description>Not sure I follow the correction by Xo Wang.

If the PLL multiplication is 2 more than the value of PLLFBD, wouldn&#039;t PLLFBD = 0×8 give a (M=8+2) 10-fold multiplication, leading to a Fosc of

8 MHz × 10 × 1⁄2 × 1⁄2 = 20 MHz?

BTW, Great tutorial!

Carl</description>
		<content:encoded><![CDATA[<p>Not sure I follow the correction by Xo Wang.</p>
<p>If the PLL multiplication is 2 more than the value of PLLFBD, wouldn&#8217;t PLLFBD = 0×8 give a (M=8+2) 10-fold multiplication, leading to a Fosc of</p>
<p>8 MHz × 10 × 1⁄2 × 1⁄2 = 20 MHz?</p>
<p>BTW, Great tutorial!</p>
<p>Carl</p>
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