The resistor would have to precisely match your current draw to get the right voltage drop. Given your predicament the easiest route might be to just burn the extra power with a 5v zener diode to lower the voltage.
What is the nominal current at 5v? 5v devices do not generally tolerate >5.5v too well.
Or throw a 7805 in there....(We learn everyday... I had to Google "Larson scanner", now I know what it is....)
Quote from: JimboZA on Feb 02, 2013, 05:33 amOr throw a 7805 in there....(We learn everyday... I had to Google "Larson scanner", now I know what it is....)A 7805 needs 2 to 3 volts differential in order to start regulating (i.e. 7 to 8 volts in minimum). It won't work for this application.There's no reason why a 5 volt board won't work at 5.5.... but if it's really necessary to throw away that 1/2 volt, why not just use a series 1N4001 diode and drop 0.7v off the end of the wall wart?The real reason that the OP's board is going nuts is not the extra 1/2 volt, but probably something else like maybe lousy filtering in the wall wart (AC ripple?).
He said the wall wart is putting out over 6v not 5.5.
Quote from: smeezekitty on Feb 02, 2013, 07:05 amHe said the wall wart is putting out over 6v not 5.5.Right you are. I gotta get to sleep... I'm seeing things. Where on earth did I get 5.5 from???
Now I know I could of used and 5v regulator. But it's to late. So I was wondering if I could use a resistor to lower the voltage. Is so can someone show me how to figure it out.
The lesson here is that a wall-wart is unlikely to be suitable for powering digital electronics unlessused with a proper voltage regulator, or unless it is already "fully voltage regulated" internally - thegood news is that mains _USB_ wall warts are _required_ to be fully regulated to 5V - this is what youshould have gone for I think.
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