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# Resistors are too warm Answered

Hello,
I am building a multi-touch surface using the FTIR method and after the surface is powered for about 10 minutes there is a heated smell in the room, I have noticed that the resistors are very warm and that a couple have started to brown or burn. I really need this table to be able to be safely powered at least 18 hours a day.

Here are my specs:

LED's:
I am using 88 Infrared LED's (T 1 3/4 INFRARED LED, http://www.allelectronics.com/make-a-store/category/340250/LEDs/Infrared/1.html )
--8 LED's per series
--forward voltage of 1.5v
--forward current of 100mA

Resistors:
I am using 2, 10 ohm 1/4 watt resistors parallel (so 5 ohms) with each of the 11 series'

Power Supply:
The power supply output's 12 volts DC and 1.2 amps

When I calculate using http://led.linear1.org/led.wiz it says I only need 1ohm 1/4 watt resistors, I haven't tried this yet but it doesn't seem to me that this would rectify the situation. My guess was that I need 1 watt resistors, but the calculator didn't seem to agree.

Any thoughts as to what my problem is? is it safe to go with 1 ohm 1/4 watt resistors? or do I need 1 watt resistors? Any help would be very appreciated, thank you!

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## 25 Replies

11010010110 (author)2008-10-18

use higher power resistors

you can split the power to multiple resistors like this

say you need 10 ohm 0.8 watt

connect 4 X 2.5 ohm resistors in series

you get 4 X 2.5 ohm = 10 ohm and 4 X 1/4 = 1 watt

if you use different resistance reistors the larger ones (in ohms) get most of the power - it does not spread evenly

to keep resistors cool dont load them above 1/2 of their watt

to keep resistors warm but not hot dont load them above 3/4 of their watt

Part A is correct, while part B isn't. You can not multiply the wattage, as ALL of the electrons are flowing through ALL of them. Your thinking of the (theoretical) properties of them both in series and parallel. But they can't be in both at the same time. Theoretically, it is possible to do it in series and then put four sets of those in parallel, but it is impossible in the modern day world. if the resistors where created with exactly the same resistance., it would work. But it will not work, because modern-era technology isn't capable of that technologically advanced processing, especially in a mass-market way.

11010010110 (author)2008-10-19

if your resulting resistor (from series connection) is the same ohm (in sum) as the original then it must dissipate the same power

for example we need 100 ohm resistor to put on 10 V
10 V / 100 ohm = 0.1 A (current thru the resistor)
0.1 A * 10 V = 1 W (power on the resistor)
and we want to use 4 X 1/4 watt resistors instead of 1 X 1 watt

series connection

lets implement it with 4 X 25 ohm 1/4 watt reistors
4 X 25 ohm = 100 ohm
10 V / 100 ohm = 0.1 A (sum resistance remains the same --> current remains the same)
now lets find the power dissipated in one of the 4 resistors
0.1 A (series circuit - current thru one is current thru all) * 25 ohm (one resistor) = 2.5 V
0.1 A * 2.5 V = 0.25 W (on one resistor)
each resistor dissipates 1/4 W

parallel connection

lets implement it with 4 X 400 ohm 1/4 watt reistors
400 ohm / 4 = 100 ohm
10 V / 100 ohm = 0.1 A (sum resistance remains the same --> current remains the same)
now lets find the power dissipated in one of the 4 resistors
10 V (parallel circuit - volt on one is volt on all) / 400 ohm = 0.025 A (crrent thru one)
0.025 A * 10 V = 0.25 W (on one resistor)
each resistor dissipates 1/4 W

2 X 2 net connection

lets implement it with 4 X 100 ohm 1/4 watt reistors
we connect 2 in series to make 200 ohm 1/2 W and then connect the 2 strings in parallel to make 100 ohm 1 W
100 ohm * 2 = 200 ohm (one series string)
200 ohm / 2 = 100 ohm (all)
10 V / 100 ohm = 0.1 A (sum resistance remains the same --> current remains the same)
now lets find the power dissipated in one of the 4 resistors
10 V (series circuit - volt on one string is volt on all) / 200 ohm (one string) = 0.05 A (crrent thru one string)
0.05 A * 100 ohm = 5 V (on one resistor)
0.05 A * 5 V = 0.25 W
each resistor dissipates 1/4 W

the resistors are inexact but the maximum possible oveload due to that is not significant. you can use more resistors and then you dont load them to the maximum

gmoon (author)2008-10-19

Parallel resistors distribute the current across each resistor, series resistors do not.

There is only one current path, so the current through each series resistors is the same. If the current at each resistor was measurably less, how could there suddenly be more current at the end?

11010010110 (author)2008-10-19

there cannot be more current the existing current divides to the number of parallel paths. in series there is one path so the current is the same. in parallel there are 4 paths so the current in each is 1/4 the sum (if the paths are the same resistance) in the parallel part where i wrote current is the same i mean the sum current thru 4 resistors

The paths aren't the same resistance. ever. There is always a very small difference in each resistor. Which is why it won't work.

11010010110 (author)2008-10-19

Padlock it does not need to be perfect to work. its not like if all the current goes thru the path that is little less resistance to see imagine some inexact resistors (with small different error in each) and try one of the examples i shown above. you should get results fairly close to what you get with exact resistors. the errors are small and not significant to an extent where it does not work gmoon the error is given in % of the resistance and not in plain ohms. with small resistors the error is small too. most resistors (last line is gold) have accuracy of +- 5 % their value and this is very accurate for allmost any purpose. say for 0.25 ohm the error is tiny unnoticable 0.0125 ohm a problem you really get with small resistors is that the resistors are of the same magnitude as everything else in the circuit. every wire and joint counts and adds resistance. you may end up with much higher resistance than what you'd expect from the resistors only

It's a fact. Electricity will find the path with the least resistance to ground. Even if the difference is in nano-ohms, it will find the one with the lesser resistance. Which is why it won't work.

11010010110 (author)2008-10-19

then explain why if you connect 2 devices to power using a power strip both work ?

Because a device can only use so much power. If there is more power to start with, then it will flow through both of them.

11010010110 (author)2008-10-19

the power a device takes is determined by the voltage it is connected to (attribute of the power source) and its resistance (attribute of the device itself) when connecting devices in parallel current (and power) always splits between them. if the source cant supply the power then the available power is divided and all appliances work on reduced power but they dont go out selectively an example is when you connect large loads (like few KW heaters) at home. the brightness of lights falls. the source cannot supply the full power required to operate both heater and lights in full power. they both operate at reduced power - not one of them goes out to leave all the power to the other what happens actually ? the voltage of the source drops when it is loaded. in the example with the heater you maybe get 245 V without and 237 V with it connected. both lights and heater work at little reduced power since they get 237 V and not 245 V. this is correct for resistors and devices that behave as resistors (heating elements and incandescent lights). devices with semiconductors / motors / arcs sometimes do go out when there is not enough power the rule you wrote is good approximation for severe stuff like short circuit etc but does not apply to resistors in general

Blah. I give up. There's only so much a 13 year old can know.

11010010110 (author)2008-10-19

one interesting tale i found once on the net was between a student and physics teacher. the teacher said that a glass of boiling water would take more time to freeze in the freezer than a glass of room temp water. the student did not agree. they decided to try. as the tale says it turned out that sometimes the boiling glass freezes first why ? when its boiling water evaporates from it. at the time when its no longer boiling a significant amount of water has evaporated. there is much less water remaining than in the other glass. now the smaller amount of water cools faster and gets to zero before the other glass with more water this is actually 3 X example the 2nd one hiding there is whether to believe the story or not. i wont give the results i got myself. i think the bet part of it is to actually try it yourself and see - like many other things in life the 3rd one is that results may depend on how exactly things are done. sure if you put just hot (and not boiling) or the freezer is too stuffed then the cold glass freezes first since there is not enough evaporation. even different amounts of water in the glasses may and probably does affect the results back to resistors. you can just sit there and calculate the thing using ohms law. but math isnt solution when it comes to principal disagreement - as here (does current flow thru the easiest path only / thru all available paths and only the amount depends on). here common sence and logic and experience rule over math speaking of common sence / logic / experience - a 'battle' like there was is one of the ways to find and study new things. it should NOT be avoided or given up. the battle has task. its not to show who is right or wrong but to find out whats really going on and let everyone study something new continue to be the way you are - it lets you study things on occasions you'd never get into would you just agree to everyone. maybe even continue right here :)

zachninme (author)2008-10-19

Now that's not true. It can be kept in mind as a pseudo-rule of thumb, but it doesn't work exactly. If it were true, resistors in parallel would take the resistance of the lower value. In reality, its the recipricol of the sums of the recipricols.

gmoon (author)2008-10-19

You're misreading my comment-- 0.25 ohms resistors are quite expensive, and generally overkill for this project. Yeah, you might be able to find one that's less than one watt, but if you're going that route, why not just choose one resistor of the correct wattage? It's much easier to find high-wattage resistors at those values than low-wattage ones.

And using 32 resistors isn't really a viable option.

most resistors (last line is gold) have accuracy of +- 5 % their value and this is very accurate for allmost any purpose. say for 0.25 ohm the error is tiny unnoticable 0.0125 ohm

Well, the effect on current is proportional, so the actual amount doesn't really mean anything. At some point, there's a sane amount of current headroom, and if you push the envelope too far, then the result is predictable failure... Actually, low-value power resistors usually have higher tolerances than 5%, anyway...

11010010110 (author)2008-10-19

its all up to you. the circuit does not mind what resistors you use (if the sum resistance is ok and power each of them gets is ok for it)

gmoon (author)2008-10-19

It's just that Padlock has a point--although I tend see more of a problem with series resistors when you're dealing with resistances < 10 ohms.

For instance, is it practical (or economical) to use four 0.25 ohm resistors to distribute the wattage? Especially since you'd need really high tolerance resistors...

I think we've all gotten away from the OP's question, though...

Your assuming that each resistor is exactly the same. They aren't. And since electricity chooses the path with the least resistance, all of the electrons will flow through the lowest value resistor, overloading it, then proceeding to the next one after that one blows.

hapticdata (author)2008-10-15

so if I were to change it into 9 parallel series' of 9 LED's that would lower it to using 1.3v and I could keep the same 5 ohm, 1/2 watt resistors and either remove the remaining 7 LED's or pair them with a 39 ohm resistor then it should work?

zachninme (author)2008-10-18

Do you have 8 more? it would probably be significantly easier to add them than to re-wire the system. They don't even have to be embedded in the board, you could just have them hanging...

gmoon (author)2008-10-15

The Forward Voltage Vs. Forward Current graph in the LED spec PDF seems to indicate a forward voltage nearer 1.4 for 100mA. If you plug in 1.42 instead of 1.5V, the resistor value is 6.8 ohms (or 8 ohms for 1.4V.)

And if your power supply drifts up by only 0.3 volts, the resistance value for 1.42 FV is 10 ohms.

Either value is 7-10 times the original resistance recommendation. That's significant.

a) Try another calculator, compare results.
b) You're not giving yourself any leeway by using the max current rating for the LEDs. Failure of the LEDs is likely if the resistors short, or the voltage drifts. Use 80 or 90 mA instead.

zachninme (author)2008-10-15

Yeah. These LEDs are both powerful & touchy... I haven't seen any LEDs like this before.

For example, reducing the voltage from 1.4V to 1.3V, the "relative radiant power" (fancy brightness?) goes down about 40%.

T3h_Muffinator (author)2008-10-15

I think you need higher wattage resistors, or shorter threads (with 2 LEDs each, probably). P = IV in watts. You're throwing 100 mA through 8 LEDs, whose accumulative voltage sums to 12V 12V * .1A = 1.2 watts, which is about 5 times larger than your 1/4 watt resistors. Try putting more resistors in parallel (beef up your wattage). The two 1/4 watt resistors you have in parallel make a 1/2 watt resistor, so just keep going till you have about 1 watt, perhaps?

I didn't do too many calculations, so I may be wrong, but I think that's your issue.

Cheers,
-Josh

zachninme (author)2008-10-15

I've been talking with Josh, trying to figure it out (we think we disproved what he just said). Besides using higher wattage resistors, you could also try adding another LED. This will lower the voltage each LED uses eversoslightly, but it will be enough for them to use significantly fewer mA. Based on the datasheet, the LEDs will be less brighter. But at 100mA, you were already "overclocking" them. Good luck with whatever you try ;-)

T3h_Muffinator (author)2008-10-15

Ignore that - I didn't calculate it right.