So this is my first instructable, hope it is usefull.

I was wondering for quite a while how could I build my own coffee heater, for the following reasons:

- I usually bring my coffe to my desktop (not the virtual one), and while I am coding or something it gets colder and colder, sometimes needing replacement (even my old Ni-Cd have a greater dutty cycle);

- Lots of power available through unused USB ports, I have 8 and use only 2 or 3 (think green);

- Usb heaters are cheap and easy to find, but I needed to build one from scratch to satisfy my ego (and impress my friends);

- I have lots and lots of scratch at home, needed to find a way to use them instead of just throwing out (think green, recycle, reuse reduce)

- and, at last but not least, I found someone who actually did it, with the math behind ( check this link ).

*This is a prototype!*- I actually built it, plugged to one of my USB and felt it going hot. The USB port still runs normally, no power or caffeine issues. But, as you'll see, it is not yet finished, no casing, no ZIF socket, no way to prevent the mug from dropping. I already have some ideas for an improved second version, comments and suggestions are welcome.

**Signing Up**

So, if I wanted to, let's say, run this at 12v safely I should have at least p=V*i (p=12*0,9=10~) 10w resistors? Or 5w each?

Just a thought. take a box of matches. throw the matches away and keep the box. cut the bottom out of the box creating a cardboard frame ( similar to making a wood frame to poor concrete in ). set it on the back of your resistor/cpu and fill to the top. Level off with a Popsicle stick or chopstick and let it harden. not only do you have a nicely insulated base, but it sits up a little higher and provides a barrier between your desk and the heat dissipation from your neat-o coffee cup warmer:)

No kidding, but does anyone know how hot does it get?

after a minute I couldn't touch it

So, what if I replace the pentium with another metal plate ? With the same resistors, the same thermal grease ?

if you use epoxy to attach

I have a project where I'm trying to generate about 350F (I don't plan to do this with a USB port, but I'm just wondering how much heat this particular setup got you...

V / I = R, so 5 / .1 (100 ma) = 50 Ohms, and 5 / 2.5 = 2 Ohms

So a 2 Ohm resistor will work IF the handshake is done, and the USB port makes the full power available. If not, I understand the USB standard just current limits it by dropping the available voltage, in which case, it won't even heat significantly. So, to make it possible to use all the power the port will provide, use a 2 Ohm resistor. Now

the formula for power is I(Sq)(Amps) X R(Ohms), so 2.5 (Sq) = 6.25 X 2= 13 Watts. Big resistor physically. And you CAN use more than one resistor, in parallel, to achieve greater power in a smaller (set of) package(s), but that requires recalculating the resistance, etc., and the math is more than I want to attempt teaching here. Use a 15 Watt 2Ohm resistor, and it should work, and WILL be safe. And 15 Watts is pretty hot. Put your hand on a 15 Watt light bulb that's been burning awhile. That's how hot the resistor will get. But it won't burn up, so it won't short, and it won't open.

Micah

thanks for the confirmation on the resistor heating problem.

so what i would do, without the hanshake (wich i know nothing about)

R = 5V / 0.5A = 10 ohm (wow thats low!)

now lets say i want to build a matrix of 4 by 4 resistors

1/10ohm = 1/R+1/R+1/R+1/R

notice how i didnt use R1 r2 r3 etc bcuz every column has to have the same values to properly distribute power

so lets see....

1/10 = 4/R

1/40 =1/R

40 =R

i must have "4" 40OHM columns

and since series resistance is Rt = R1+R2....+RN

and i want 4 of them also

40/4 =10

its 10 Ohm!... magic? no, its a mathematical "coincidence" it has something to do with the fact that its a square matrix but the procedure should be the same

now how much power will each resistor dissipate?, Pt = P1+P2+P3...+PN. i think it doesnt matter if its parallel or series im almost 90% sure of this

so P= 5V*0.5 a

P= 2.5 Watt total

2.5W / (4*4) =

0.15625 W/R... meh its too low... so umm hows the hanshake thingy goes?

either that or we try a smaller matrix... 2 by 2 maybeh... im getting my protoboard & my coffee cup back on monday now im really really interested on this project