<?xml-stylesheet type="text/xsl" href="https://community.element14.com/cfs-file/__key/system/syndication/rss.xsl" media="screen"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:slash="http://purl.org/rss/1.0/modules/slash/" xmlns:wfw="http://wellformedweb.org/CommentAPI/"><channel><title>Half the Battle of Making Low Power Devices</title><link>/members-area/personalblogs/b/blog/posts/half-the-battle-of-making-low-power-devices</link><description>Written by elecia . On episode 104 , we talked to Andreas Eieland of Atmel about choosing low power processors, comparing them with EEMBC’s benchmark ULPBench, and tips for achieving low power. If you are making a low power device (like a wearabl...</description><dc:language>en-US</dc:language><generator>Telligent Community 12</generator><item><title>RE: Half the Battle of Making Low Power Devices</title><link>https://community.element14.com/members-area/personalblogs/b/blog/posts/half-the-battle-of-making-low-power-devices</link><pubDate>Fri, 08 Jun 2018 19:37:36 GMT</pubDate><guid isPermaLink="false">93d5dcb4-84c2-446f-b2cb-99731719e767:71a68afc-ad86-4503-9e6e-6d5a90331abc</guid><dc:creator>rajesh2610</dc:creator><slash:comments>0</slash:comments><description>&lt;p&gt;Isn&amp;#39;t power = (V^2)/R ?&lt;/p&gt;&lt;img src="https://community.element14.com/aggbug?PostID=21108&amp;AppID=293&amp;AppType=Weblog&amp;ContentType=0" width="1" height="1"&gt;</description></item><item><title>RE: Half the Battle of Making Low Power Devices</title><link>https://community.element14.com/members-area/personalblogs/b/blog/posts/half-the-battle-of-making-low-power-devices</link><pubDate>Thu, 05 Nov 2015 15:09:23 GMT</pubDate><guid isPermaLink="false">93d5dcb4-84c2-446f-b2cb-99731719e767:71a68afc-ad86-4503-9e6e-6d5a90331abc</guid><dc:creator>crjeder</dc:creator><slash:comments>0</slash:comments><description>&lt;p&gt;The power column seems wrong. Isn&amp;#39;t the power disipated by the resistor I=40mA times the voltage drop at the resistor? &lt;span&gt;The current is allways 40 mA since this is a series circuit (as &lt;span&gt;[mention:b0bc65b9ecdc4307bd967592f00e340a:e9ed411860ed4f2ba0265705b8793d05]&lt;/span&gt; saied)&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&amp;nbsp;&lt;/p&gt;&lt;p&gt;I. e. for the first line 40 mA * 0,04 V = 1,6 mW&lt;/p&gt;&lt;img src="https://community.element14.com/aggbug?PostID=21108&amp;AppID=293&amp;AppType=Weblog&amp;ContentType=0" width="1" height="1"&gt;</description></item><item><title>RE: Half the Battle of Making Low Power Devices</title><link>https://community.element14.com/members-area/personalblogs/b/blog/posts/half-the-battle-of-making-low-power-devices</link><pubDate>Tue, 16 Jun 2015 17:11:20 GMT</pubDate><guid isPermaLink="false">93d5dcb4-84c2-446f-b2cb-99731719e767:71a68afc-ad86-4503-9e6e-6d5a90331abc</guid><dc:creator>shabaz</dc:creator><slash:comments>2</slash:comments><description>&lt;p&gt;I think there are some numerical errors in this post, that don&amp;#39;t make sense (like in the table, how 40v can be measured when the applied voltage was 5v), and 1 V across a 10 ohm resistor means 100mA through the resistor, not 10mA. I&amp;#39;m on a mobile device so did not read the article fully though.&lt;/p&gt;&lt;img src="https://community.element14.com/aggbug?PostID=21108&amp;AppID=293&amp;AppType=Weblog&amp;ContentType=0" width="1" height="1"&gt;</description></item></channel></rss>