DIUTC: Two questions about MOSFETs and resistors - Page 1
Forum section: electronics / beginners

author: TomS_
author_rank: Frequent Contributor
timestamp: June 29, 2016, 09:23:28 am
content: Do I Understand This Correctly.Hello everyone. I have two question, one about calculating dissipation in a MOSFET to work out if I need a heatsink, and the other about resistors.I know this is really basic stuff, and I think I have it right, so prepare your clue-by-fours anyway...First off, resistors.I want to drop the voltage across a rail to turn on an LED if power is present. Sounds simple right? Lets say I have two rails, one is +5V the other is +48V.Take the attached image (resistors.png). If the working voltage of the LED is 2.4V, and I have a 5V supply, Ive calculated:R = (VS- VLED) over ILEDassuming 20mA for ILEDthis gives(5 - 2.4) / 20 = 130ohm or 150ohm as next best available valueFor the 48V supply:(48 - 2.4) / 20 = 2280ohm or 2.7K as next best available valueThe 48V situation is pretty severe in terms of wattage dissipated, which I calculated as 48 * 0.02 = 0.96 or roughly 1W, so a pretty "hefty" resistor needed there.Is that math sound??Now for the MOSFET.I want to switch on/off a load that draws a maximum of 1.5A. Im using a Fairchild FDC855N as an example, datasheet here:https://www.fairchildsemi.com/datasheets/FD/FDC855N.pdfNow take the second attached diagram (mosfet.png). This is a logic level MOSFET (which I only learned existed yesterday!) which has an RDSof 0.0036ohm at 4.5V gate voltage. I also learned to fudge this a bit, so lets call it 0.00432ohm.The formula is R * I2to give watts dissipated, so with this I figure:0.00432 * 1.52is 0.00432 * 2.25 = 0.00972 is 9.7mWConvert this in to temperature rise:62.5 * 0.00972 = 0.6075 + 25C ambient = 25.6CThis doesnt seem right to me. Switching 1.5A continuous would only cause a 0.6 degree rise above ambient? Just seems low... I suppose Im trying to work out if I should be adding any kind of heat sinking capability to this little package, or how much, which from the datasheet seems to be mostly via the drain pins connected to a large copper area? But if the dissipation is that low, perhaps not very much is needed?Any pointers or have I actually got it?Cheers

author: danadak
author_rank: Super Contributor
timestamp: June 29, 2016, 09:50:10 am
content: Math is correct. One question -62.5 * 0.00972 = 0.6075 + 25C ambient = 25.6CWhere is .00972 coming from, is that the 2 oz Cu thermal R ?If the switching of MOSFET occurs at a very low rate then your Pdiss calculationsare correct. However if you are driving at some higher frequency and or duty cycleyou have to take into account dissipation due to gate drive and duty cycle. This might help -http://electronicdesign.com/boards/calculate-dissipation-mosfets-high-power-suppliesRegards, Dana.

author: TomS_
author_rank: Frequent Contributor
timestamp: June 29, 2016, 10:52:35 am
content: Hi Dana,0.00972 was the result of the RDSohms value multiplied by the square of the current to be switched. Nothing to do with the copper thickness.Per the datasheet, this was 0.0036ohm at 4.5V, I fudged it a bit to 0.00432ohm (which is ~20% extra), hence:0.00432 * 1.52= 0.00972mW62.5 being the degrees per watt dissipated by the MOSFET, so 62.5 * 0.00972 gives me 0.6 degree temperature rise for 1.5A. Or maybe that should be produced...?My application wont involve any high frequency switching, just on (mostly) and occasionally off for brief periods of time (couple of seconds.) Essentially it would be to "reboot" a logic circuit.

author: ebclr
author_rank: Super Contributor
timestamp: June 29, 2016, 11:37:08 am
content: You math is perfect but some details can shade the real world situation20 ma is the maximum current on the regular led, if you use 10 ma the Led still be bright on a good level ( of course less than with 20 ma), and the power dissipation on resistor will decrease 4 times, you don't need to use the maximum current, except if you really need the maximum possible bright

author: Unknown
author_rank: Unknown
timestamp: June 29, 2016, 12:17:07 pm
content: Hi,Just a quck comment about your maths. When writing out your values try and keep your units the same as in the equation, you have volts and milliamps together. You obviously know what you meant because your answer is correct, but doing that might bite you in the backside one day especially if it's something more complex.For your 48V case, you can get LEDs which produce their rated brightness with low currents (like 2mA) which will make the resistor dissipation much less of an issue.With the MOSFET, once it is fully turned on then the channel resistance is as quoted and so the dissipation is low. As long as it is switched on and off quickly, there will be no problems.

author: matseng
author_rank: Frequent Contributor
timestamp: June 29, 2016, 12:26:36 pm
content: Quote from: TomS_ on June 29, 2016, 09:23:28 am0.0036ohm is 3.6 mOhm. The datasheet says 36 mOhm, so you're off by a factor of 10 at the resistance. But I guess the temp rise will be rather small even with this resistance...

author: ebclr
author_rank: Super Contributor
timestamp: June 29, 2016, 12:42:55 pm
content: Regarding mosfetYou assume the mosfet goes to the low resistance instantly, but this is not true he needs some time, and is on this time where all the power lose will happen , who will reflect on a power dissipation and heating

author: TomS_
author_rank: Frequent Contributor
timestamp: June 29, 2016, 02:53:16 pm
content: Quote from: matseng on June 29, 2016, 12:26:36 pmOops.I would like to say that was a test, but right you are, off by a factor of 10.So that means a 6 degree rise in temperature, which still doesnt sound too bad as you suggest.

author: Audioguru
author_rank: Super Contributor
timestamp: June 29, 2016, 04:30:53 pm
content: When the Mosfet is turned on, it has the resistance of a short piece of wire. Does a short piece of wire get hot with your 1.5A of current in it?