I have an Uno and a Mega, and I decided to measure the voltage on the "3.3V" and "5V" power pins. On both boards, the "3.3V" power pins report 3.3V. But the "5V" power pins give strange readings: on the Uno, the "5V" reports 3.3V (as if it was the "3.3V" power pin), on the Mega I get 3.99V. Should I not be getting roughly 5V in both cases?
have you got anything else connected ?
and i assume you are using a multimeter ?
What are you using to power the board?
If you feed 5v (e.g., usb) into Vin, it's going to drop on the way to 5v from the linear regulator.
dochawk:
If you feed 5v (e.g., usb) into Vin, it's going to drop on the way to 5v from the linear regulator.
Thanks, I've just found this out. I had a 5V power adapter going into the power jack. I need minimum 7V into the power jack or 5V directly to the 5V pin. With 7.5V in the power jack, I get 4.9V now on the 5V pin.
A small note on 5V on the 5V pin, DON'T connect the USB cable to a PC when you do. You'll backfeed 5V to the PC. Most modern PC's are smart enough to disable the USB port but it's tempting luck. Does not apply for Pro Mini and Nano as they have a diode in line with the USB power.
element222:
Thanks, I've just found this out. I had a 5V power adapter going into the power jack. I need minimum 7V into the power jack or 5V directly to the 5V pin.
Important point here - forget the "power jack". Yes, you can power it with 7 - or 9V if you only have a UNO or Mega and nothing else connected to it but I really suspect you want to connect other things.
If you do, at some stage you will be waning to use more than a few tens of milliamps whether from the output pins or the magic "5V"terminal and you are going to wonder why it no longer works.
Save yourself the pain. Feed 5 V into the 5 V pin - or the USB jack, and you are set.
septillion:
A small note on 5V on the 5V pin, DON'T connect the USB cable to a PC when you do. You'll backfeed 5V to the PC. Most modern PC's are smart enough to disable the USB port but it's tempting luck. Does not apply for Pro Mini and Nano as they have a diode in line with the USB power.
The point against this being likely however, is that many or most powered USB hubs do in fact, connect the accessory 5 V supply rated at something like 2.5 A directly to the 5 V terminal on all outputs and the input because without the accessory 5 V supply, the input must feed those outputs and you do not want to insert any device - such as a diode - which will cause a voltage drop.
So given the relative number of Arduino UNOs (some even genuine) and powered USB hubs, I feel if this were a frequent problem, it would be announced to a far greater audience than this forum!
There is absolutely no suggestion that this can damage the UNO itself; transfer of power beyond 500 mA in either direction might trip the polyfuse.
I'm not worried about damaging Arduino's, they are dirt cheap. I am however worried about my computer....
Hmm . . . thinking aloud . . . could the user simply short Vin to +5 if using an adequately regulated +5 input? Or does Vin feed anything other than the crummy linear regulator?
Paul__B:
If you do, at some stage you will be waning to use more than a few tens of milliamps
I plan to power just one servo from the 5V pin. I think Arduino can handle this (with 7V in the power jack). As far as preventing 5V flowing back into the PC, I found this: how to use a diode to prevent it:
septillion:
I'm not worried about damaging Arduino's, they are dirt cheap
Yes, but there are other considerations, like labor and time. At this stage of my project, my Arduino is now buried deep in a tight enclosure, beneath other components. It would cost time and labor to replace it. I'll go with the safer 7V / power jack route because of this.
I plan to power just one servo from the 5V pin. I think Arduino can handle this
No. Servos, being motors, draw lots of (too much) current for that really.
element222:
5V directly to the 5V pin
I'd advise against doing that. From the UNO schematic, you can see that the correct way to power from 5V is to do it via the USB connector.
If you do as you say, someone could apply voltage to the power jack and there'll be a conflict between the two power sources.