MOSFET Datasheet Confusion - Page 1
Forum section: electronics / projects

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: April 26, 2025, 10:28:57 am
content: Hello all, hope everyone is doing well.A tiny bit of background, I am building a capacitive discharge spot welder. The MOSFETs I am using are IRL3803 (datasheet below). Probably not a great choice but I can get them for cheap (0.35USD/each)Since I am designing a spot welder, my load is a pulsed load and the steady state thermal resistance calculation is not very relevant. I am hence using this graph given in the datasheetIts pretty intuitive but the Y axis is the transient thermal impedance, junction-to-case. I am more interested in the junction-to-ambient resistance, so that I may calculate the junction temperature after the pulsed load, and make sure I have enough MOSFETs in parallel to survive the welding process.Is it a matter of using these values to figure out what junction to case corresponds to what junction to ambient temperature?Like here 40/0.75 = 53.33, which would mean 1C rise in case temp means a rise of 53.33C on the junction.I'm pretty sure I'm missing something here.

author: tooki
author_rank: Super Contributor
timestamp: April 26, 2025, 12:50:06 pm
content: You are missing something: you’ve invented a ratio where there is none. Thermal resistances function like resistors in series.So that means that the junction-to-ambient thermal resistanceincludesthe junction-to-case thermal resistance.So 40-0.75 = case-to-ambient thermal resistance of 39.25 °C/W.The idea is that the junction-to-ambient number gives you an idea of how it behaves when using it without a heatsink, while the junction-to-case number is used for making your calculations for a heatsink.For example, if you have a little heatsink with a thermal resistance of 25 °C/W (heatsink-to-ambient), and the layer of thermal paste adds 0.5 °C/W (case-to-heatsink), then your total junction-to-ambient thermal resistance is:(junction-to-case) + (case-to-heatsink) + (heatsink-to-ambient) = 0.75 + 0.5 + 25 = 26.25 °C/WSuppose you dissipate 5W, and that your device’s ambient temperature (remember, “ambient” means the immediate surroundings of the device and heatsink, so if it’s inside a case, then “ambient” means the temperature of the air inside the case) is 35°C.Then your junction temperature will rise to: 35 + (5 x 26.25) = 166.25°C, within spitting distance of the 175°C datasheet limit.What it also means is that no matter how huge a heatsink you use, with the best thermal grease known to man, your junction temperature will always rise by at least 0.75 °C/W. So assuming an ambient temperature of 25°C, then the maximum allowable junction temperature rise is 175-25=150°C.This means your absolute maximum power dissipation is150°C/(0.75°C/W) = 200 W(And that’s exactly what the datasheet says, too, for 25°C ambient.)That’s assuming a theoretical cooling system (heatsink or whatever) with zero thermal resistance, which of course doesn’t exist. And this demonstrates how datasheet values cannot be taken at face value, since they’re often theoretical fantasy numbers that cannot be achieved in reality.For fun, assume we kept it submerged in liquid nitrogen, which is -196°C. Then our maximal junction temperature rise is 175 - (-196) = 371°C.Assuming a hypothetical case-to-ambient resistance of zero (since I have no idea what the thermal resistance of boiling liquid nitrogen is), then our maximum dissipation would be:371°C/(0.75°C/W) = 494.667 W

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: April 26, 2025, 01:28:27 pm
content: Many thanks for the comprehensive answer. This clears up everything!

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: April 26, 2025, 01:32:19 pm
content: I wonder if i can treat the 40C/W as 40Cs/J (since J=Ws) and use that for some exact milliseconds accurate thermal limitWell probably should learn to walk before running

author: mtwieg
author_rank: Frequent Contributor
timestamp: April 26, 2025, 01:38:00 pm
content: So first of all, the Rja number in the datasheet is not very meaningful. For ICs this number is usually based on measurements with the IC mounted on a small board which is then suspended in the center of an empty chamber with still air (see JEDEC standard JESD51-2, though this is just one example, not all devices follow this standard). Usually this ends up being a worst-case result. If your PCB is larger, has large copper pours, or has other means of transferring heat away, then your actual Rja can be much lower.As for your pulsed power scenario, if you have a good heatsink for the case, then one can make the assumption that the case temperature changes very slowly, and therefore you only need to consider average power for calculating case temperature (Tc = Rca * Pavg). Though good luck figuring out what Rca actually is...The junction to case temperature drop is where the transient curves for Rjc are useful. Knowing your peak pulse power and duration and duty cycle, it tells you the Rjc, so you can calculate the peak Tjc = Rjc * Ppulse.Then you can calculate the peak junction temperature Tj = Tc + Tjc.This is a pretty rough approximation, but it's usually good enough so long as the case is bonded directly to the heatsink/spreader. Unfortunately for you, the D2PAK does not lend itself well to this, because the "case" is effectively the drain tab soldered to the PCB (putting a heatsink on the plastic case is not effective).Quote from: BadeBhaiya on April 26, 2025, 01:32:19 pmNope.Do you have an idea of what your power dissipation will look like (peak/duration/duty cycle)? Maybe the Rja of 40C/W is already sufficient.

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: April 26, 2025, 01:54:13 pm
content: Quote from: mtwieg on April 26, 2025, 01:38:00 pmHmm this is something that i have only estimated using some assumption.I calculated it to be 70W/MOSFET (10 in parallel), at a total current is 1000A. Of course this depends on a few factors such as the quality of caps that i use. I estimated the on time to be around 100us but that's too short for spot welding (i found this out later). Also I'm considering switching losses to be negligible since it's just one pulse every several seconds, but i still have to figure out the proper turn on time. Plus on top of that I'm assuming perfect turning on time for all MOSFETs.Its way too early to know a few things so I'm wondering if i should just make a board and then play around

author: PCB.Wiz
author_rank: Super Contributor
timestamp: April 26, 2025, 08:13:50 pm
content: Which package ?For this you need all the thermal mass you can practically get, so TO220’s and bulk heat sink.You also need serious spike clamping, and you will consume a lot of fets..

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: April 26, 2025, 08:22:30 pm
content: These FETs are D2PAKI'm almost certain i should get 3x the amount i need because I'm doing to be killing quite a few of them (i sure hope not though)

author: PCB.Wiz
author_rank: Super Contributor
timestamp: April 26, 2025, 09:05:00 pm
content: Quote from: BadeBhaiya on April 26, 2025, 08:22:30 pmSwapping out a damaged fet will be funMaybe a small carrier PCB that can be bolted to a heatsink, and includes TVS per FET is easier to service?PCBs are cheap, you can get creative with BUS BARS and stackups.

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: April 27, 2025, 07:56:41 am
content: My hot air station is too underpowered so I'll probably use a plumbers heating gun

author: mtwieg
author_rank: Frequent Contributor
timestamp: April 27, 2025, 12:57:49 pm
content: Quote from: BadeBhaiya on April 26, 2025, 01:54:13 pmNo harm in doing some calculations as an exercise. So if pulse duration is 100us and pulses happen every second, that's such a low duty cycle you can basically consider single-pulse behavior. So for 100us, the Zjc would be about 0.041 C/W. Power during the pulse would be about 90W, so Tjc would be 3.7C, basically negligible. With a pulse duration of 1ms, Zjc would be 0.15 C/W, Tjc would be 13.5C, still quite low.Average power dissipation will be in the milliwatts, so the 40C/W Rja of the D2PAK might be sufficient with no additional heatsink.So it doesn't look like thermals are a big deal for the MOSFETs. Should consider protection against voltage spikes, that might be the bigger issue...

author: BadeBhaiya
author_rank: Regular Contributor
timestamp: April 28, 2025, 09:35:27 am
content: Quote from: mtwieg on April 27, 2025, 12:57:49 pmAny idea what I should be looking out for? I'm honestly a bit clueless when it comes to power electronics in general, much less about switching high current loadsI'm thinking some beefy TVS diodes across the FETs and the welding electrodes is probably a good idea? Also should have a choke on the MCU power rail to avoid latchups? Besides that I'm really not sureAlso attaching the schematics. 3rd page is the MOSFET subcircuit. Its copied for all the 10 FETs

author: tooki
author_rank: Super Contributor
timestamp: April 28, 2025, 05:27:13 pm
content: Quote from: PCB.Wiz on April 26, 2025, 08:13:50 pmWouldn’t TO-247 be even better of the THT packages??

author: PCB.Wiz
author_rank: Super Contributor
timestamp: April 28, 2025, 07:26:58 pm
content: Quote from: tooki on April 28, 2025, 05:27:13 pmYes, but that given part number has a TO220 choice.