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[Reply #30 on:]

SXR is just my jet ski. it's a 2005 Kawasaki SXR 800.

I'm starting DeVry on Jan 13th, getting my bachelor's in EE, so I'll be learning all of this stuff over the next few years

Check out www.thebestcasescenario.com, my username there is the same as it is here. I've got some cool projects going on over there, adn some cool ones that are already finished up

[Reply #31 on:]

one thing I'm noticing is that I'm already using 3 analog inputs for the voltage reading. will I need one for each current reading or two? if one then I need 6 analog pins, which is fine (0-5), but if I need two then I'll need 9 pins. can I use some of the digital outs as inputs in the same manner if need be?  :-/

[Reply #32 on:]

No, only the 6 analog pins go into an internal multiplexer for the A/D reading.
You can add some relays to select what gets connected, or an external analog multplexer chip to select what you are connecting also.
Depending on how you install the shunt will determine whether you need to take a reading on one side of it (so maybe you have one side at ground) or two (it is in the middle of something else, and you need to take a reading from both side and then find the difference between the two).
What are you connecting to that needs the 3 ranges you have selected?

[Reply #33 on:]

this is the project I want to make this for:
http://www.thebestcasescenario.com/forum/showthread.php?t=25018

and this is the LCD voltmeter I've already got working 100%:
http://www.thebestcasescenario.com/forum/showthread.php?t=25092

I'm using a standard 400W computer power supply as a benchtop supply unit. I'm going to have a ground, 3.3v, 5v, and 12 supply lines. I've got the LCD set up to read and display the voltages out of each line, and I want to be able to push a button and cycle between voltage, amperage and wattage screens on the LCD to aid in my electronics testing/designing.

It's probably not needed, but it'd be nice to see just how many watts or amps my project is using, and exactly how much voltage the supply is putting out

[Reply #34 on:]

Ok, read this too.

http://www.allaboutcircuits.com/vol_1/chpt_8/4.html

[Reply #35 on:]

thanks a ton for that! I'll have a read through it when I get home from work tonight

[Reply #36 on:]

System seems to have dropped a post I made.
You'll want the shunt at the source.
You are already reading the source itself, if you add your remaining 3 lines to the other side of the the shunt then you can measure the voltage on both sides.  Then
(Vsource - Vload)/Shunt Resistance is the current draw.
Power (watts) = IV, or doing some substituting with V=IR you can get P= IIR, or P=VV/R, take your pick to calculate it.