Timeline for Arduino Uno R3: Directly supply regulated 5V to 5V pin?
Current License: CC BY-SA 3.0
14 events
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| Jun 11, 2020 at 15:10 | history | edited | CommunityBot |
Commonmark migration
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| Apr 28, 2020 at 9:12 | comment | added | gwideman | @Bozzy With the MOSFET off, current can still flow through the MOSFET's diode from USBVCC to +5V. However, that current will be tiny, because in this scenario (>6.6V at VIN) we have +5V supplied from regulator U1 to the +5V line. Thus there will be no (or very little) voltage drop across that diode, hence very little current. The diode would need a forward drop of say at least 0.4V and more like 0.6V for appreciable current through it. | |
| Apr 27, 2020 at 7:57 | comment | added | Bozzy | @gwideman sorry but I don't understand: if the diode in the mosfet permits current from USBVCC to +5V, how could U5A+mosfet disable USBVCC as an input? | |
| May 6, 2016 at 4:02 | comment | added | gwideman | @ProBackup The diode in the MOSFET is always present and never "disabled" per se. However, when the MOSFET is in its ON state, the drain-source path has such low resistance that the diode that parallels it is irrelevant. (So the diode is only relevant when the MOSFET is OFF, and in that case permits current only in the direction from USBVCC to +5V.) | |
| S Feb 9, 2016 at 10:05 | history | suggested | Pro Backup | CC BY-SA 3.0 |
upper/lower case made more consistent, abbr. v expanded to volts when not meant as a PCB label, PCB labels maked in backtick escapes
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| Feb 9, 2016 at 9:27 | review | Suggested edits | |||
| S Feb 9, 2016 at 10:05 | |||||
| Feb 9, 2016 at 1:58 | history | edited | Passerby | CC BY-SA 3.0 |
added 54 characters in body
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| S Feb 9, 2016 at 1:53 | history | suggested | Pro Backup | CC BY-SA 3.0 |
Trying to make the text more readable, by inserting headings, und put more emphasis on the 2 warnings
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| Feb 8, 2016 at 22:54 | comment | added | Pro Backup | @Passerby You answer explains in good dummy language what is going on. However I don't quite understand what you would like to say with "Diode on the mosfet, is a Body Diode". My best guess is something like: Although the Mosfet contains a diode that normally would protect currents from flowing into the direction of the USB host, this internal diode (protection) is disabled as soon as voltage is applied to Arduino its "USBVCC". | |
| Feb 8, 2016 at 22:46 | review | Suggested edits | |||
| S Feb 9, 2016 at 1:53 | |||||
| Apr 26, 2014 at 13:02 | comment | added | gwideman | But I do concur that connecting a 5V supply to the Arduino shield +5V will work, but as Passerby emphasizes, the user has to be sure not to connect that supply AND USB at the same time. A possible result of connecting both would be to damage either the USB host, the external supply, or preferably, blowing of fuse F1. | |
| Apr 26, 2014 at 12:58 | comment | added | gwideman | "If VIN is detected, and higher than 3.3v, the opamp drives the line low": Actually there's a voltage divider between VIN and CMP input to U5A, which means that VIN has to be higher than 6.6V to disable USBVCC as an input. Further, since there's a diode between the power input jack PWRIN and VIN, (adding say 0.6V drop) and NCP1117's dropout is 1 to 1.2V, that means the external supply should be above 7.2V to shut off USBVCC, even though without USBVCC the external supply could supply reliable power down to 6.6 to 6.8V | |
| Apr 26, 2014 at 12:48 | comment | added | gwideman | Regarding the ICSP (and ISP) headers: The pin that's connected to the +5V net is ISP VTG, which is intended for the target device (here Arduino) to supply power to the programmer, which allows the programmer to accommodate to +5V or +3.3 targets. It's not intended as an input for 5V, so it's not evidence in favor of connecting a 5V supply to the +5V net. (Though not strongly against either.) | |
| Apr 15, 2013 at 8:08 | history | answered | Passerby | CC BY-SA 3.0 |