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<?xml-stylesheet type="text/xsl" href="https://devzone.nordicsemi.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/" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>nRF52 current consumption in relation to supply voltage</title><link>https://devzone.nordicsemi.com/f/nordic-q-a/30320/nrf52-current-consumption-in-relation-to-supply-voltage</link><description>Hello, 
 in the Online Power Profiler , the current consumption increases, when the voltage is reduced. What is the reason for this? 
 Isn&amp;#39;t there any documentation on the power consumption in relation to the supply voltage, with LDO or DCDC?</description><dc:language>en-US</dc:language><generator>Telligent Community 13</generator><lastBuildDate>Wed, 07 Feb 2018 17:33:24 GMT</lastBuildDate><atom:link rel="self" type="application/rss+xml" href="https://devzone.nordicsemi.com/f/nordic-q-a/30320/nrf52-current-consumption-in-relation-to-supply-voltage" /><item><title>RE: nRF52 current consumption in relation to supply voltage</title><link>https://devzone.nordicsemi.com/thread/120158?ContentTypeID=1</link><pubDate>Wed, 07 Feb 2018 17:33:24 GMT</pubDate><guid isPermaLink="false">137ad170-7792-4731-bb38-c0d22fbe4515:d5eb5f43-390f-4a96-9082-7fa7c2f5c8e5</guid><dc:creator>AmbystomaLabs</dc:creator><description>&lt;p&gt;It just basic ohm&amp;#39;s law. When you utilize a dc/dc converter it is simply converting power at one voltage to power at another voltage.&lt;/p&gt;
&lt;p&gt;Since power = volts x amps then when you reduce the input voltage the current must increase for the power output to be the same.&lt;/p&gt;
&lt;p&gt;Or you can show the relation as (Vout/Vin) x I out = I in.&amp;nbsp;&amp;nbsp; So the ratio of output voltage to input voltage times the output current is the input current.&lt;/p&gt;
&lt;p&gt;There are some minor changes to this due to the dc/dc regulator efficiency. But this is best to get the concept across.&lt;/p&gt;
&lt;p&gt;With an LDO it is a different story since amps out is equal to amps in. No matter what the input voltage is the input current is always the output current.&amp;nbsp; This is why an LDO is very inefficient since the difference in power between input and output is given off as heat in the LDO.&lt;/p&gt;&lt;div style="clear:both;"&gt;&lt;/div&gt;</description></item></channel></rss>