SMD Packages, Continuous Current, Junction Temperature and Heat Dissipation. - Page 1
Forum section: electronics / beginners

author: IvanIlievMCU
author_rank: Newbie
timestamp: July 11, 2020, 02:04:06 pm
content: Hi everyone,I'm working on 4 Layer / 1 oz copper plating PCB that's supposed to operate on 5V and draw between 0.5A - 1.5A. It will be powered with anything between ~5V - 12V DC (6V Recommended) and be able to withstand up to 36V.I was planning to use the MAX17561 (36V - Max / I(LIM) - 4.2A) to supervise the input power and let it run through an LDO (TL1963A-Q1 / (20V Max / 1.5A)) to step it down to 5V if it's in the range of ~5V - 12V.Here is what i don't understand.The datasheet of the MAX17561https://datasheets.maximintegrated.com/en/ds/MAX17561-MAX17563.pdfsays that the absolute maximum allowed junction temperature is +150C and includes the following note "Note 2: Junction temperature greater than +125°C degrades operating lifetimes". Just below it is the package information that shows a Junction to ambient temperature increase of +39C per Watt on a four layer board.Knowing all of that and doing the calculations tells me that i can't use the MAX17561 even if i supply it with the absolute minimum operating voltage of 4.5V noted in the datasheet let alone 6V -12V as i was planning on using, and 1A (Even though the datasheet says the programmable current limit is anywhere between 0.7A to 4.2A)1A x 4.5V = 4.5W4.5W x 39C/W = 175.5CI'm getting similar results when i do the calculations for the LDO, but it should be able to withstand the input power since it's a much bigger package with better thermal characteristics and a thermal pad underneath it, heatsinks for it are also available.It should be noted that the board, with the default configuration coming out of the fab will draw no more than 500-800mA, but the continuous current draw could increase up to 1A, with relatively long periods of it drawing 1.5A depending on the user configuration. The PCB has to operate / withstand a wide input voltage range and be relatively fool-proof because it's going to be used in remote locations, away from repair shops by not the most tech savviest people.How do i go about it, are my calculations wrong, and if they aren't, are there any solutions for a problem like that.

author: Rick Law
author_rank: Super Contributor
timestamp: July 11, 2020, 08:22:00 pm
content: First, I am no expert. I am sharing where I think you erred hoping/expecting that some real expert will either confirm me or correct my error. We can both learn something.I think you are calculating power wrong here.Quote from: IvanIlievMCU on July 11, 2020, 02:04:06 pmHere, you are calculating the wattage you are supplying, but not the wattage the MAX17561 has to dissipate.It is the wattage the chip has to dissipate that heats the chip. So we have to evaluatehow many watts the chip is consuming, so we must find the voltage drop across the chip. I can't find the Voltage Drop or Voltage Forward listed in the electrical characteristics, but I did find the resistance.Obviously Vdrop(Vforward) will be missing because it depends on your Vmaxlimit. When you are supplying a voltage below between selected Vmaxand Vmin, I take it you expect the same voltage out (or as much as it can).On page 1 of the spec under "Benefits and Features", listed is "Low RON100mΩ (typ)", so when ON, the thing looks like a 100mΩ resister, typically. You are supplyingat allowable min, but say it turns on just fine (instead of locking out as I suspect it may do as well with the smallest of voltage sag)...P=I*V, and V=I*R, so P=I*I*RP = 1A * 1A * 100mΩ = 1A * 1A * 0.1Ω =0.1 WattThe 4.5W you calculated is what you supplied and expect the chip passed along to your load, 0.1 Watt is what the chip consumed.When you are lock-out, then you use the undervolt/overvolt leakage current and (100nA in both cases) and do the P=VI to see how many watts it has to dissipate. Since current is in mere 100 nano Amp, you would be talking nano-watts to micro-watts, not even milli-watts.EDIT:added and then removed a situation when Voltage in > selected max pretending it would be a Vdrop(heat dissipation had it regulate the voltage down) calculation -- it is pointless since the chip is suppose to lock-out instead of regulate it down. The paragraph was replaced with lock-out current calculation instead.Additional EDIT - initial reply I had said "below allowable limit" with your 4.5V supply, wrong wording, you are at limit, not below, so reworded accordingly.

author: IvanIlievMCU
author_rank: Newbie
timestamp: July 11, 2020, 09:35:47 pm
content: Thanks, you have a point about the voltage supervisor actually and i really appreciate your effort to look through the datasheet for me, but what about the LDO, i've attached a screenshot of the formula used to calculate the maximum junction temperature.TI gave an example with an input of 6V, output of 3V3, Continuous Current of 500mA and ambient temperature of 50C:P = 500mA * (6V - 3.3V) + 10mA * (6V) = 1.41Using a KTT package, the thermal resistance is in the range of 23°C/W to 33°C/W,depending on the copper area.So the junction temperature rise above ambient is approximately equal to:1.41W × 28°C/W= 39.5°CThe maximum junction temperature is then equal to the maximum junction-temperature rise above ambient plus the maximum ambient temperature:TJMAX= 50°C + 39.5°C = 89.5°CFollowing the same formula supplying just 1/2 of the maximum allowed voltage and 2/3 of the current gives me:P = 1A * (12V - 5V) ....... (Even ignoring I(GND))P = 1A * 7P = 7W7W*28°C/W = 196°CTJMAX= 50°C + 196°C = 246°CHow does a device like the Arduino manage have an input voltage range up to 12V or more while drawing 500mA without a problem, or a Raspberry PI which draws multiple Amps while using much much smaller SMD LDO / Power distribution packages.I've also only recently got into electronics and PCB design, is there even an advantage of using such a big package like the TO-263-5. Scrolling through Mouser and DigiKey i can find much smaller LDOs with similar or better ratings but none of their thermal characteristics make sense to me.What's the point of having a tiny LDO "capable" of delivering up to 1A / 5V with an input voltage range between 6-10V or so for example if it's going to seemingly melt by passing half a Watt through it

author: IvanIlievMCU
author_rank: Newbie
timestamp: July 11, 2020, 09:48:10 pm
content: The Datasheet of the MAX17561 shows a MIN input voltage of 4.5. Why do you it would lock up if i supply it with with >5V and program the undervoltage threshhold to 4.5V

author: Rick Law
author_rank: Super Contributor
timestamp: July 11, 2020, 09:56:19 pm
content: Quote from: IvanIlievMCU on July 11, 2020, 09:48:10 pmQuick one first, I had a wording error - I know you are at limit, my mind thinks any jitter or voltage sag you are out of luck. And I wording it that way, it was wrong.I reedited but apparently this is right before/after the edit.

author: Rick Law
author_rank: Super Contributor
timestamp: July 11, 2020, 10:14:13 pm
content: Quote from: IvanIlievMCU on July 11, 2020, 09:35:47 pmI replied to the later one first because it was basically a one-line reply. Now replying to this one since this requires a longer reply:That calculation looks like it is for a voltage regulator - regulating 500mA@6V down to 500mA@3.3V, while the chip itself consumes 10mA at 6V.When you are regulating from a 6V supply down to 3.3V, it does so by burning that extra voltage away as heat. That is what the above formula shows and that is what a linear voltage regulator does, LDO or otherwise.The MAX17651 chip spec doesn't read like it regulates the voltage - it reads like it is a limiter. If your input voltage is within the range you select, then it turns ON. If it is outside that range, then it turns OFF.It does not try to turn the 6V down to 3.3V and give 3.3V out to you. It just turn OFF and give you nothing but some leaks. So, the only wattage in question is the leaked current when it is OFF.