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Jun 11, 2020 at 15:10 history edited CommunityBot
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Aug 11, 2016 at 22:31 comment added Gábor Móczik Chip designing is totally out of my profession. 1V 10mA consumption equals to 10mW dissipation. If you make ANY type of regulator with 10% of effieciency you will end up at 90mW PSU + 10mW sensor dissipation. Isn't that too much? I mean, this will be a single-chip temperature sensor, won't that dissipation affect the measurement significantly?
Jan 10, 2013 at 17:32 comment added placeholder you should be made aware of this new answer to a similar sort of question here electronics.stackexchange.com/questions/53587/…
Jan 7, 2013 at 6:05 vote accept alan
Jan 7, 2013 at 6:05
Jan 7, 2013 at 2:08 answer added Olin Lathrop timeline score: 1
Jan 7, 2013 at 1:06 comment added alan @OlinLathrop I edit my question again, and thinks different methods for some time, hope you can check
Jan 6, 2013 at 16:52 history edited alan CC BY-SA 3.0
added 2062 characters in body; edited tags; edited title
Jan 4, 2013 at 12:18 vote accept alan
Jan 4, 2013 at 12:18
Jan 4, 2013 at 10:10 answer added Anindo Ghosh timeline score: 3
Jan 2, 2013 at 19:30 review Close votes
Jan 4, 2013 at 19:53
Jan 2, 2013 at 18:45 answer added placeholder timeline score: 5
Jan 2, 2013 at 17:12 comment added Olin Lathrop Charge pumps can certainly step down as well as up. At high step down ratios, they look more like current sources. The current is a function of the capacitances and the pumping voltage and frequency. Capacitive charge pumps are common in un-isolated supplies that only need to put out a few mA. They are also fairly efficient. You would likely need to add some sort of regulation afterwards. This can be as simple as a zener shunt regulator since from 220V to 3V the charge pump will look like a current source.
Jan 2, 2013 at 15:33 comment added Olin Lathrop If the "connector" is just to plug into the wall outlet, then I don't see what the problem is. Use whatever the standard wall plug is for wherever this device will be used. As for 220 VAC on a chip, not gonna happen. The supply needs to be external to the chip, then whatever chips you have run from the resulting low voltage DC. A charge pump may be suitable, as I suggested to your original question days ago.
Jan 2, 2013 at 15:13 history edited alan CC BY-SA 3.0
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Jan 2, 2013 at 14:16 comment added Olin Lathrop We STILL don't know whether this power supply needs to be isolated. That must be decided first since the solutions either way will be very different. You say this will power a sensor, but where does the output of the sensor go? You also ask about a connector, so it implies the sensor signal may go elsewhere. If the sensor signal is just to be displayed locally, for example, then the supply could be isolated, but then the question about connector makes no sense. Also, what is this about, the electronics or the connector. Pick one.
Jan 2, 2013 at 14:09 review First posts
Jan 2, 2013 at 15:39
Jan 2, 2013 at 14:09 history reopened W5VO
Jan 2, 2013 at 13:14 history edited alan CC BY-SA 3.0
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Jan 2, 2013 at 13:02 history edited alan CC BY-SA 3.0
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S Dec 31, 2012 at 22:09 history suggested DrFriedParts CC BY-SA 3.0
Rewrite + personal letter in an attempt to salvage a potentially valuable future member
Dec 31, 2012 at 21:37 review Suggested edits
S Dec 31, 2012 at 22:09
Dec 31, 2012 at 19:11 history closed Nick Alexeev
Dave Tweed
Leon Heller
Olin Lathrop
The Photon
not a real question
Dec 31, 2012 at 13:22 answer added Russell McMahon timeline score: 5
Dec 31, 2012 at 11:09 answer added user16324 timeline score: 2
Dec 31, 2012 at 4:43 review Close votes
Dec 31, 2012 at 19:14
Dec 31, 2012 at 4:39 answer added placeholder timeline score: 8
Dec 31, 2012 at 4:07 history edited alan CC BY-SA 3.0
added 120 characters in body; edited title
Dec 31, 2012 at 3:02 history edited alan CC BY-SA 3.0
added 10 characters in body; edited title
Dec 31, 2012 at 2:46 answer added DrFriedParts timeline score: 6
Dec 31, 2012 at 2:39 history asked alan CC BY-SA 3.0