need to know basic equation for solving power dissipation??? - Page 1
Forum section: electronics / beginners

author: doctorm
author_rank: Regular Contributor
timestamp: May 25, 2010, 02:03:45 am
content: so im studying electronics as a hobby and im hitting a snag in my book.it gets done talking about power and how to solve for it (im in the first part of the book about basics) and then it goes off too how power is dissipated.one example is how a 60 watt bulb can dissipate 1 watt-hour in like secs or something like that.so is there a basic equation that i can reference to help solve this and other examples it gives?if you need the page im reading i can scan it for you to see what im talking about...

author: EEVblog
author_rank: Unknown
timestamp: May 25, 2010, 02:35:23 am
content: A reference will help. Sound like one of those silly physics types questions instead of a real practical electronics question (the light bulb is the giveaway!)But basically there is no magic formula which is going to help you with thermal design and heat dissipation. It varies greatly with surface area, surface type, enclosure size and ventilation, free air or moving air etc.The standard technique is to do some basic dissipation calculations and then add in large safety factors. Or often simply build a prototype and measure.Dave.

author: allanw
author_rank: Frequent Contributor
timestamp: May 25, 2010, 03:21:05 am
content: The problem you're doing just sounds like unit conversions. For these types of problems you don't really use a formula, just use the definition of what a watt is and such.60 W = 60 J/s.A watt-hour has units of J/s * 3600s = 3600J.

author: Polossatik
author_rank: Frequent Contributor
timestamp: May 25, 2010, 05:48:47 pm
content: Like Dave say's (in one of his blog's also) calculating heat dissipation is an Art in itself.The thing they most likly try to say is that if switch on an incandescent light bulb then it can dissipate a LOT of heat before it's warmed up (in a room temperature env), simply because heat "needs to go somewhere" and if the glass is cool it can heat the glass, after that it needs to "transfer" all the heat from the glass to the air, and air is pretty hard thing to heat up (any stuff that is used to isolate houses is basically air packed in small bubbles) and that is also not a bad thing, seen an Incandescent light bulb generates the most heat when just starting up (the resistance of the filament rises when heated).The one thing you need to remember in electronics is basically that heat (and not current or volt itself) is the killing factor.If you put like 10 volt and 1 amp trough a 10 ohm resistor, than that resistor needs to dissipate 10 watt, if that is a "10 watt" resistor then it will (barely) survive.However you can perfectly put like 10volt and 10 amp trough that same resistor ( = 100 watt) if the signal / current only flows for like 1/10 the of a second and then 9/10 of second there is no current.It then "evens out" on 10 watt.In practice you don't want to be so close to the watt specification of that resistor of course, more like only use about max 80% of the rating.bottom line, when starting and then messing around with 5v circuits who blink leds or so, don't worry to much about heat yet.for things where you need cooling (big transistors in Audio amp, poser supply etc) just look what other similar circuits recommend/use and use something that is at least that big. Most of the time you'll start with an existing circuit anyway.

author: Unknown
author_rank: Unknown
timestamp: May 25, 2010, 05:55:12 pm
content: Yes, thermal calculations are often very complex, but if it's in a basic chapter in an electronics textbook, after they just told you about voltage, current and resistance, I think allenw's answer is likely to be close. I wouldn't expect them to require you to come up with a thermal model, more likely conservation of energy (60W in a bulb is 60W out, in either light or heat). If you assume a 0% efficiency (not too far of the mark for an incandescent lamp), it would dissipate 60W, or 60W * (1/60)h = 1Wh in one second.

author: Polossatik
author_rank: Frequent Contributor
timestamp: May 25, 2010, 06:06:36 pm
content: Quote from: alm on May 25, 2010, 05:55:12 pmMinute?

author: Unknown
author_rank: Unknown
timestamp: May 25, 2010, 06:27:55 pm
content: You're correct, it should obviously have been (1/3600)h for one second, or (1/60)h for one minute

author: tecman
author_rank: Frequent Contributor
timestamp: May 25, 2010, 10:46:30 pm
content: I think a basic starting point is power dissipation. Any electronic device, or circuit, has a power in and a power out. The power out may be electrical (power supply) or mechanical (motor) or other forms of energy (like your light bulb example). The light bulb is a tricky one since it is hard to measure the output in light and heat (IR). Other devices are easier, like a power supply. A supply might draw 1 amp at 120 volts, or 120 watts. The output is, as an example, 60 volts at 1 amp, or 60 watts. So the supply is dissipating 60 watts (of heat).Power dissipation in electronic components, like diodes, transistors, resistors and even capacitors, is generally a calculable value. This dissipation is nearly always in the form of heat and must be taken into account in design, component selection and thermal design (heat sinking, air flow, etc).Paul

author: Pyr0Beast
author_rank: Contributor
timestamp: May 26, 2010, 01:23:35 am
content: polossatikQuoteAnd on-off cycles or heat-up cool-down, which cause all sorts of mechanical stresses and failures.

author: Polossatik
author_rank: Frequent Contributor
timestamp: May 26, 2010, 02:44:32 pm
content: Quote from: Pyr0Beast on May 26, 2010, 01:23:35 amYep indeed, the main reason why for example a lamp that is switched on/off often (hallway or so) cr*ps out "faster" than the lamp in the living room that burns for hours/day...

author: Mechatrommer
author_rank: Super Contributor
timestamp: May 26, 2010, 04:13:32 pm
content: P = VI, Pt = VIt = E, Eff(%) X E = Light Energy... (100 - Eff(%)) X E = Heat Energy.P,V,I,t = Electrical Power, Volt, Current, timeE = Electrical EnergyEff(%) = Device Efficiency in %Light Energy =Heat Energy = refer to some Thermodynamic TextConservation of Energy = refer to Elementary Physics

author: kc1980
author_rank: Regular Contributor
timestamp: May 26, 2010, 10:19:13 pm
content: Most incandescent light bulbs radiate far more infrared radiation than visible light.Correct me if I'm wrong, but isn't the dissipated power = voltage drop x current. Some of the generated heat is conducted away through the light bulb structure. I'd say that most of it is dissipated as IR radiation.I guess I don't understand the original question....

author: Pyr0Beast
author_rank: Contributor
timestamp: May 27, 2010, 09:48:05 pm
content: Light bulb will dissipate everything that goes in, same with everything else.60W input = 60W outputIt is just usable energy (light) is only a fraction of what goes in, but it all converts to heat non the less.

author: kc1980
author_rank: Regular Contributor
timestamp: May 28, 2010, 08:36:10 pm
content: Oh, keep in mind that Lumens are based on the response of the human eye. Wikipedia sums it up well: "The lumen (symbol: lm) is the SI unit of luminous flux, a measure of the power of light perceived by the human eye. Luminous flux differs from radiant flux, the measure of the total power of light emitted, in that luminous flux is adjusted to reflect the varying sensitivity of the human eye to different wavelengths of light."There is a difference between photometry and radiometry. The latter is what we are interested in.