Power Dissipation - Page 1
Forum section: electronics / beginners

author: JaseG
author_rank: Contributor
timestamp: November 08, 2016, 08:58:36 pm
content: Hi,Just after a little clarity on something, although I think I'm on the right track.I'm making a constant current circuit (based on the on on the eevblog) which uses a Mosfet and a 1 ohm resistor and I want to ensure that the chosen Mosfet will be within it's operating range. The circuit currently produces 2v after the Mosfet (at the source pin) and I'm assuming that any extra voltage present before the mosfet (at the drain pin) needs to be dissipated as heat.So if my understanding is correct, with a 5v input voltage I'll need to drop 3v and at 2 amps that's 6watts of power that needs to be dissipated.This sounds pretty simple. Maybe there's more to it, but for a simple ballpark figure is this good enough?ThanksJase

author: tatus1969
author_rank: Super Contributor
timestamp: November 08, 2016, 09:06:01 pm
content: it's that simple

author: JaseG
author_rank: Contributor
timestamp: November 08, 2016, 09:29:08 pm
content: Thanks!I've seen some other calculations that used the Rds(on) value from the data sheet as well. Should I add this in as well??P = I2x Rds(on)P = 4 x 0.047P = 0.188Although small in comparison to the wattages from before, it would be nice to know.

author: JaseG
author_rank: Contributor
timestamp: November 08, 2016, 10:02:26 pm
content: Actually, thinking about it more, I probably don't need to do this.From what I understand, the Rds(on)value is the resistance in the fet when fully switched on. I guess this is used to calculate the voltage drop across the component.My case seems different as I'm using an opamp to control the voltage at the source pin.

author: Zbig
author_rank: Frequent Contributor
timestamp: November 08, 2016, 10:07:55 pm
content: Rds(on) figure tells you the lowest possible resistance you could achieve with the given MOSFET fully-on when used as a switch. Obviously not relevant in your case as your intention is to actually dissipate considerable power, i.e. use the device in its linear (non-saturated) region.EDIT: You got it by yourself as I was writing this

author: Cerebus
author_rank: Super Contributor
timestamp: November 08, 2016, 10:32:56 pm
content: Find the datasheet for your MOSFET and look for the SOA or Safe Operating Area curve in the graphs. That shows the combinations of voltage and current that the device can handle. As the name suggests, if you are operating the device at a point that is inside the SOA then all's good. The SOA curve gives you a much better idea what you can push through a device than the 'headline' power, voltage and current limits for the device.After this, your next job is to figure out that your heatsinking is adequate for the dissipation that your device is throwing out.

author: JaseG
author_rank: Contributor
timestamp: November 08, 2016, 10:49:56 pm
content: Thanks Zbig - good to have confirmation.@Cerebus - From what I can see I'm well below the curve here, thanks for pointing it out. The headline figures have been a bit misleading in the past, but being able to find the important info doesn't seem to be too difficult. It's just not obvious for a newcomer.Regarding the heatsinking, I need to re-watch another clip on the eevblog which covers the topic quite thoroughly. I have some small heatsinks from ebay that don't have any data, would it be acceptable to dissipate a few known wattages through the heatsink and take some measurements? From this I'd imagine it wouldn't be too hard to reverse engineer a best guess at what the thermal resistance of the heatsink is.

author: Cerebus
author_rank: Super Contributor
timestamp: November 08, 2016, 11:49:20 pm
content: Quote from: JaseG on November 08, 2016, 10:49:56 pmI've done that before myself for heatsinks with uncertain pedigrees. The results can be a bit hit and miss so I'd advise being conservative in using the results.The aluminium cased wire-wound power resistors make a useful heat source for this. Just use dab of thermal paste and run off a bench power supply. Better still are TO-220 cased power resistors but most people don't have those lying around and they tend to be a bit expensive to pick up just for this.

author: Audioguru
author_rank: Super Contributor
timestamp: November 09, 2016, 01:03:09 am
content: You can look at heatsinks on the website of as heatsink manufacturer like Aavid and pick one that has the same dimensions as the junk one you bought from ebay. Its spec's will say its thermal resistance.

author: tatus1969
author_rank: Super Contributor
timestamp: November 09, 2016, 05:54:56 am
content: put your transistor into that 6w dissipation, add the heatsink and watch temperature develop. i have an infrared thermometer for that, these 5$ models at ebay do surprisingly well. watch the temperature develop. when using infrared you need to stick some black tape on the heatsink if it is not black itself.question is the max temp that you want to allow. if it is a hobby project you can go to the max, lets say 120°c at the heatsink. for a product you need to take the max ambient temp into account that you want to allow.

author: JaseG
author_rank: Contributor
timestamp: November 09, 2016, 08:35:58 pm
content: Quote from: tatus1969 on November 09, 2016, 05:54:56 amYep - This is what I was thinking of doing. I have the components ready to go so it seems like the most logical solution. It will also give me the most accurate data for my application. A different package could have different thermal transfer properties.Quote from: tatus1969 on November 09, 2016, 05:54:56 amI'll need to check the datasheet for the mosfet, it's probably only rated to 125 at the junction. I have a small quantity of mosfets and heatsinks so I could parallel them, and I salvaged a fan from a pc power supply so I can get some airflow as well.I did some rough calculations last night and depending on the design specs I allow, I might have to dissipate 20w.I'd like it

author: tatus1969
author_rank: Super Contributor
timestamp: November 09, 2016, 08:48:50 pm
content: Quote from: JaseG on November 09, 2016, 08:35:58 pmYou can easily calculate the junction temp when knowing the dissipated power, and all involved thermal resistances: deltaT = P * RthRth would be the sum of Rth[junction-case] and Rth[case-heatsink]. The latter depends on the thermal coupling between transistor and heat sink. For a TO220 and thermal grease, maybe 0.1 ~ 0.5 K/W.But as always: staying away from the limits helps staying alive