I've been visiting Instructables once in a while, and I realized it was time to re-start building stuff. I used to unmount-mod my "toys" when I was a kid - teenager (like blowing out a little train and putting its mottor in a GI-Joe like helicopter to spin its blades, cool), and at some point of my life I forgot how amusing that was.

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.

Step 1: The physics behind

IMPORTANT: I can not be held responsible for any damage caused to USB ports, computer parts, power supplies, coffee mugs, the coffee itself or greenhouse effect. I checked the hole thing for the maximum safety and built mine at my own risks, I expect you do the same.

Some basics on Ohm's Law:

I = E / R

Which means, the current through the resistors (I) in Amperes is equal to the volts applied (E) divided by the resistance (R). Consider E=5 (USB power), R=14 (resistors used):

I = 5 / 14 = 0.357143

So, we're spending around 350mA, safe enough for most USB ports with a 500mA limit.
Now, to check how much power will be sent to our tasty coffee:

P = E x I

We already know the values of I and E, so:

P = 5 x 0.357143 = 1.785714

meaning we get close to 1.8 watts, enough to keep it hot till the last drop. You can calculate your own to give more/less heat, but keep in mind the limits (5V, 500mA for USB ports).

'Some considerations:'

- First things first. The resistors should be able to handle the power they will be dissipating; Thou shall not use resistors smaller than 1 watt or thou shall have unpredictable and unpleasant surprises... I took mine from old ATX power supplies, judging by theyr size they're at least 1w for sure (the body is almost 1 inch long). Guess you'll not spend a dolar to get them on local stores.

- Using 2w or 4w resistors (or bigger) will give you more safety, not heat. Trust me when I say this is technology, not magic.

- You can use a different number of resistors, as long as you keep the final resistance inside the calculated values. A smaller resistance could burn your USB, cause personal injuries or several other situations involving your hot coffee.

Kids, don't try it at home without reading everything carefully!
The resistors are somewhat critical parts, you should stick to the values calculated (by me or you). When you connect two (or more) resistors in series, the final resistance value is increased by summing all their values. So, 6.8+6.8=13.6 (the tolerance of each resistor may take it a bit higher or lower).
Connecting them in parallel DECREASES the final resistance value, drastically, so don't do that, unless you really know all the formulas.

ATTENTION: (part 2 - thanks RetroPlayer)
If you connect this project to a non-powered hub (meaning no AC/DC adapter plugged in), the 500mA from the USB is actually split by the number of ports, so this probably wouldn't work on that type of hub, unless you use the other ports for unpowerable purposes (like connecting digital cameras, scanners, printers and other stuff that doesn't depend os USB power to work). Anyway, a powered hub will ensure 500mA to each port, therefore being a better idea in case you really need to use a hub (or eventually would forget and plug all your tiny USB stuff).

- Ensure your favorite mug can handle extra heat, you will not be pleased if it melts over your stuff. The same goes for the casing if you build one. By the way, good ideas for casing are wood or some thick plastic, I believe metal should be avoided because it could dissipate ALL the heat and/or eventually cause a short circuit.
<p>Can I use any plaque of metal instead of the pentium processor ??</p>
<p>Why you are using the &quot;Clock&quot; ?</p>
<p>so you just place the connected resistors on the processor? is there any need for it to be soldered into the pins on the processor.?</p>
<p>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? </p>
<p>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:)</p>
<p>No kidding, but does anyone know how hot does it get?</p>
<p>after a minute I couldn't touch it</p>
This is great
To be short, the only thing that produces heat is the two resistors, right ? And the Pentium is only there to let the heat flow through a large panel, in order to heat something, am I right ? <br> <br>So, what if I replace the pentium with another metal plate ? With the same resistors, the same thermal grease ?
I love this. It always gets so cold where I live that anything that involves <a href="http://www.trirom.ca/en/" rel="nofollow">heating in Kamloops</a> and isn't outrageously expensive is a good thing.
This is so neat! I love the idea of using something like this for <a href="http://www.thegentlemenplumberscalgary.com/en/heating.html" rel="nofollow">heating in my Calgary</a> office! Sounds like fun, I may have to try and make this!
<br>if you use epoxy to attach
I realize this is an old post, but what kind of temperatures did you get with this setup?<br><br>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...
Even though it's smaller, this works well with my old 90Mhz 486 processors
im making this with my i486 dx processor i pulled from a really old ibm computer.
so... is it the processor or the wire between the resistors that is heating up?
How hot does it get
near 3 million degrees
mine went up to 20 million degrees celcius. <br />
Easy reading - very helpful, bravo man!
somebody with an overclocked cpu put their motherboard in oil and played some game (I think it was WOW) for like, 2 hours. they used it to fry potatoes.
its nice, thanks for the idea for our thesis
so....if i used something like copper wire and did a squiggily thing....kind of like a stove top where it is spiraled.....wouldnt that work the same if it was in direct contact of the coffee cup? I really dont want to waste money (even if a few cents) on a processer.....
what about thiscally steking the heater to the bottom and jattaching a usb port and putting a case on so all you do is plug in th usb to the cup
Hey i wonder if 2x 8ohm resistor be ok ?
it would not get as hot but it will work
as long as the w is correct
Ok thanks.
if you plan to use other usb devices depending on the configuration there might not be enough power for it all.
no problem
lol. i took some magnet wire (like 20 ft) and i put it in a spiral. i found it had around 1k ohms (multimeter) (this was some thin wire) so i hooked it upto 3 volts. and it works pretty good.
uhh, wouldnt it make much more sense to make one with a peltier unit, which is actually meant for heating and cooling. dont get me wrong, this is a great idea and really original, i would never have thought of making a heater of of an old processor, but using a peltier would probably make it heat better.
about how hot does this get?
Nice instructable!, Simple yet effective!, i had been wanting to know the MAX current rating of a USB port. im going to try this one :), i have all the materials within my room... err somewhere in this mess however, a resistor with a higher power rating would result in less heat transfer to the proccessor, wouldnt it? since a higher power rating means its designed to dissipate heat more effectively. i think heat loss around the resistor (the part not directly in contact with the proccessor ) would be increased as power ratings for the resistors increase, id try to use the lowest rating possible, always considering that heat has to be transfered entirely to the proccessor "cooling" the resistors as much as possible. wow... i did learn something at the "heat transfer" course! LOL
Well, thanks a lot. I was so concerned about USB safety that I didn't think about that (actually, I have never used a resistor for heating purposes). I still don't know much about heat transfer, but all the time I kept thinking what could happen if used a smaller power rating resistor (I mean, wouldn't it possibly burn at some point? Maybe cause a short and damage USB?) I wanna test these possibilities on the 'Mark 2' :)
What i would do, if i wanted to make those tests would be to use a 5V power supply, i made one at school, if you dont have one maybe a buddy of yours has one (or you can make one ), then plug in the circuit and monitor the current on it, that way you can avoid damaging the USB port, since a power supply can handle more current, i dont think youll get a short circuit from a burnt resistor (im not 100% sure tho) , a smaller resistor will be more likely to burn , but thats where the thermal paste and the processor come in... so yeah, basically try testing it with a power supply
Nice tip about the power supply, thanks. I believe the most common is that resistors get 'open' when they burn, just like a fuse, but I've seen cases when they got 'short circuited'.
The common cycle is: Step 1 - heat up some, but not excessively, Step 2 - approach excessive heat, Step 3 - enter "thermal runaway" (where current rapidly goes to max, and component burning occurs, Step 4 - Short circuit, short duration, as this is where heating REALLY hits the ramp, Step 5 - Smoke, burn, open, no more current. Its during the thermal runaway portion that other components are most likely to fry. I speak as a "fry baby", from personal experience. And 35 years in component repair, replacement, troubleshooting, and test. Fried a bunch, fixed less. Micah
Thanks for sharing this information. It is good to know from who really saw it happening from, you know, a (long) road (man, I really mean it, 35 years is older than me). By the way, do you know how safe would it be the watts rate for our resistors to heat the coffee without damaging the circuit itself or the USB's?
If you are familiar with Ohm's Law, it allows you to calculate the current (I or Amps), the Voltage (V) and the resistance (Ohms or R), and from them, the power the resistor needs to dissipate. Since USB is 5 Volts, and USB current gives you 100 ma, unmodified, or 2500 ma, after appropriate handshaking, you can use the formula as follows<br/>V / I = R, so 5 / .1 (100 ma) = 50 Ohms, and 5 / 2.5 = 2 Ohms<br/>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<br/>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.<br/>Micah<br/>
where did you get the 100 ma figure? gb78 had said that it was 500 ma, i haven't tested this yet but the power rating on resistors only means the power the resistor is capable to dissipate, in other words, a resistor rated 15W will stay relatively cool under a 15 watt load, again im looking for confirmation on this cuz i haven't run any tests. also, yeah it would probably be better to make sets of resistors to make a better distribution of the heat source on the microprocessor.
I read somewhere else, probably on the CR4 Engineering board I read and write on, that the standards for USB include an initial insertion of the device, at up to a 100 ma signal load, used for establishing between the device and the USB controller, that the voltage and current are within specs. After the initial contact, which I understand takes usually less than 2 seconds (about how long it takes before the activity lights on my 4 port hubs come on when I insert a device, actually), the port will offer up to 500 ma. But I understand from the same discussion that the USB standard is only a recommendation, and that most USB controller designers save money by putting a current limiter in the circuit at 500 ma, and leaving out the handshaking entirely. Thus, if you use the 500 ma for the power calculation, since the voltage is a known (it is set at 5 volts, + or - .5 volts), the current controls the maximum power the circuit will deliver to your resistors. And the truth, from practical experience, is that when you use a 15 watt resistor, and dump 15 watts of power across it, it isn't just a warm resistor. One company that makes them calls them its "SandOhm" line of resistors (or used to, maybe I date myself, but my experience with component electronics started in 1960, and continues through today), and they made those out of a sand and glue concoction. I burned an imprint of the body of one into my finger once, while touch testing it. And it was NOT overloaded. For your use, the formula for two resistors in parallel is (R1 X R2) / (R1 +R2). I am not certain any longer, but I believe for three resistors it is (R1 X R2 X R3) / (R1 +R2 +R3) so that for N resistors it becomes (R1 X ... RN) / R1 + ... RN). And if you are comfortable with the algebraic manipulations, you can clearly see that no number of parallel resistors will ever have a total resistance equal to the smallest resistors. BTW, I think this formula only works for equal resistance values, but I can't remember for sure. You'll have to do more research to be certain, and I would appreciate the feedback if you do. I am always up for a refresher course.
you are sort of right with the calculations but that formula i think only works for 2 resistors in parallel, the basic fromula is 1/Rt = (1/R1)+(1/R2)+(1/R3)+... +(1/Rn)<br/>thanks for the confirmation on the resistor heating problem.<br/>so what i would do, without the hanshake (wich i know nothing about) <br/>R = 5V / 0.5A = 10 ohm (wow thats low!) <br/>now lets say i want to build a matrix of 4 by 4 resistors<br/>1/10ohm = 1/R+1/R+1/R+1/R<br/>notice how i didnt use R1 r2 r3 etc bcuz every column has to have the same values to properly distribute power<br/>so lets see....<br/>1/10 = 4/R<br/>1/40 =1/R<br/>40 =R<br/>i must have &quot;4&quot; 40OHM columns<br/>and since series resistance is Rt = R1+R2....+RN<br/>and i want 4 of them also<br/>40/4 =10<br/>its 10 Ohm!... magic? no, its a mathematical &quot;coincidence&quot; it has something to do with the fact that its a square matrix but the procedure should be the same<br/><br/>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<br/>so P= 5V*0.5 a<br/>P= 2.5 Watt total <br/>2.5W / (4*4) =<br/>0.15625 W/R... meh its too low... so umm hows the hanshake thingy goes?<br/>either that or we try a smaller matrix... 2 by 2 maybeh... im getting my protoboard &amp; my coffee cup back on monday now im really really interested on this project<br/>
Hmm. Your calculation method beats mine. I knew it once, too, but seldom used more than two resistors in parallel, so I always had a hard time remembering that formula. So, you are absolutely right. Mine was limited to two in parallel, and not more, and yours covers an infinite number. I never got the quick and right (both) answer to the complex resistance circuit problems in school, either. Good thing in the US Navy I was mostly dealing with designs that used only one, or at most two. Or else I didn't have to calculate them, since I was using well documented schematics to repair and troubleshoot from.
Ok, first of all without the handshake you should never get .5A directly off the USB port simply with a 10 ohm resistor. The port should simply not provide that much current, as it should be limited to 100ma. As for spreading things out over a matrix you will get more surface area but less heat off of each resistor. Too many resistors and I'll bet you won't be able to measure the difference between the "hot" resistors and ambient temperature.
i see, well with 0.1A @ 5 V you get at most 0.5W power, we cant do much heat distribution if any at all I know If you make a big matrix heat will be lower on each resistor, i just did the math to show that, i was looking for some sort of point in between total loss of heat and just having one hotspot... But you are right in that it might be pointless because it all depends on the exact use of the heater... i mean if your cup has a nice flat bottom 1 hotspot will do a nice job, however if you want to keep the entire area at the same temperature (I kno this isnt possible, bear with me) it would be nice to have more resistors... LOL i just realized how much discussion there is over this instructable, Its so cool, the fact that a USB Coffee cup heater can generate so many ideas is a good thing XD instructables.com rocks!
Actually I was saying that as you approach the maximum rating of a component often the amount of heat produced to compensate will increase non-linearly. This is generally not a good thing, but if you want to produce heat intentionally then I guess it can be. Furthermore, "heating something up" and "keeping it from dropping to ambient temperature as quickly" can be different things. With this design I somewhat doubt you will be able to do much more than delay the cooling of the cup slightly. If you were to accumulate energy over time, possibly increasing the voltage as well, then use a more proper electrical heating element in waves, I think you would have a better chance of keeping your coffee hot. Using a boost-converter or a charge-pump off the USB port isn't something I would personally like to do, but its probably the only way to get anything out of it. Then once you are fully charged, pump all the energy you can out into a proper heating element. Better housing, insulation, and directed heat flow would perhaps help but that's assuming you are producing enough heat to do anything to begin with.
Here, I did part of your homework for you: <a rel="nofollow" href="http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=LM78S40CN-ND">http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail?name=LM78S40CN-ND</a> . Hook that up, use some op-amps and a micro controller, and grab a variety of resistance wire called Nichrome wire (eg: <a rel="nofollow" href="http://www.skycraftsurplus.com/index.asp?PageAction=VIEWPROD&amp;ProdID=820">http://www.skycraftsurplus.com/index.asp?PageAction=VIEWPROD&amp;ProdID=820</a> or <a rel="nofollow" href="http://wardsci.com/product.asp?pn=IG0008178&amp;sid=2008FS&amp;eid=2008FS&amp;mr:trackingCode=BE5C6261-4C9E-DD11-AA1A-000423C27502&amp;mr:referralID=NA).">http://wardsci.com/product.asp?pn=IG0008178&amp;sid=2008FS&amp;eid=2008FS&amp;mr:trackingCode=BE5C6261-4C9E-DD11-AA1A-000423C27502&amp;mr:referralID=NA).</a> Note on the first Nichrome site I linked it states the heat the wire will achieve at different current ratings and its resistance per foot. That's pretty impressive, and if you built up a good 40V in a cap and then just pumped it all out into a coil of this stuff you could probably get a reasonable amount of heat going.<br/>
15 watts can be VERY hot. My first soldering iron was 17 watts.
"Fried a bunch, fixed less." I know all about that one. My skill levels improving now though, maybe as much as 25% sucsess!

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