L
ltj
Guest
If you had a small 3.3V, 50mA circuit how could you make it to accept a
5-24VDC input range? Cost is an issue.
5-24VDC input range? Cost is an issue.
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Are you talking about responding to logic signals or measuring analogIf you had a small 3.3V, 50mA circuit how could you make it to accept a
5-24VDC input range? Cost is an issue.
ltj wrote:
If you had a small 3.3V, 50mA circuit how could you make it to accept a
5-24VDC input range? Cost is an issue.
Are you talking about responding to logic signals or measuring analog
signals?
--
John Popelish
A switching regulator converts one voltage to another with minimal"John Popelish" <jpopelish@rica.net> wrote in message
news:40AAD8B4.EC2096A8@rica.net...
ltj wrote:
If you had a small 3.3V, 50mA circuit how could you make it to accept a
5-24VDC input range? Cost is an issue.
Are you talking about responding to logic signals or measuring analog
signals?
oops...for powering the circuit. A regulator will get too hot if I drop too
much voltage across it, so what is a good way to have a wide power input
range?
A switching regulator converts one voltage to another with minimal
power loss. Here is a tutorial on the subject to get you started.
http://www.national.com/appinfo/power/files/f5.pdf
This page on the National Semiconductor site will design a switching
regulator for you, to your specifications:
http://www.national.com/appinfo/power/
That said, 50 ma drawn from a 24 volt source represents only 1.2
watts, total. So a linear regulator would have to waste no more than
(24-3.3)*.05=1.03 watts. That is about what a square inch of heat
sink could dump.
--
John Popelish