Hello.
I cut off the cable of an old broken dc adapter. On the end of the cable, where cut the adapter off, I connected a battery catridge and inserted batteries.
Everything is working just fine. But I measured the voltage on the barrel plug...and it's only 8.02 V.
Can I damage the arduino with this less power? Or shall I really try to get at least 9V? I know it's maybe a stupid question but I don't want to damage my arduino or the lcd display and the other hardware.
The Arduino spec is 5VDC to 12VDC, so running it at a voltage toward the lower end of that range simply means that the voltage regulator on your Arduino won't get as hot. Just make sure that you're feeding it DC, and not AC (the wall-wart should have something on it saying what its output is).
The risk I see is the battery/charger system you describe. I wouldn't recommend having batteries in the holder(s) AND running off the AC adapter at the same time unless you've taken the time to include a battery-charging circuit that matches the adapter output to the battery pack type and voltage.
Yes, that is fine in concept. Maybe not implementation.
It is hard to tell, but from the looks of the photo you have made a 6V pack and a 3V pack sharing a ground and then having two V+ terminals (one 3V & one 6V). Only the 6V looks connected to the barrel plug. ??
If in reality the two V+ terminals are connected together, then you are "charging" the 3V pack from the 6V pack. Ungoodness will result.
This is what you need:
red ----------- barrel plug center
|
2 cell, 3V pack
|
black
|
red
|
4 cell, 6v pack
|
black---------- barrel plug outer
I can't imagine how you read 8V, unless you are wired the way I suggest but your cells are old, or one is bad.
Thank you for your post.
They are not normal batteries, they are rechargeable batteries. So at least in Germany, they do not output 1.5V, but 1.2V. Well at least the text on the batteries says 1.2V. But I measured values between 1.28V and 1.34V when I checked all 6 batteries.
I have wired them as you suggested. It's right you can't see it well on the photo.
But If I understood the first reply right, I could also use only 4 batteries à ~1.3V - resulting in 5.2V?