LDO overheating - Page 1
Forum section: electronics / general

author: DiTBho
author_rank: Super Contributor
timestamp: July 25, 2023, 01:59:33 pm
content: Found a Arnewsh board during a dumpster diving, and it looks ~ok.According to its board user manual the powersupply should be 7.5VSo, there are two LDO units in parallelLT1086-CT3.3 -------> 7.5V to 3.3V (this one overheats!!!)LT1086-CT5.0 -------> 7.5V to 5.0Vand in my opinion, the LT1086-CT3.3 has too much Vdrop (7.5 - 3.3V)Watt = I_load * VdropI'm surprised the board was designed this way, as hack, what If1) I dissolder the LT1086-CT5.02) I shortcircuit Vin and Vout of the LT1086-CT5.03) So I can use an external regulated 5V PSU instead of the 7.5V onedoesn't look better?sbc5307-board-psu.png(14.38 kB. 466x408 - viewed 1150 times.)

author: DiTBho
author_rank: Super Contributor
timestamp: July 25, 2023, 02:06:11 pm
content: (like other saved boards, this one also has no dBUG on its flash, instead there is a strange bootloader. But that's a second problem, first you need to be sure that the hardware doesn't burn out)

author: Scrts
author_rank: Frequent Contributor
timestamp: July 25, 2023, 08:22:38 pm
content: Does J6 feed some secondary board that needs 7.5V? If not, you can power it with 5V and do what you've planned to do.

author: DiTBho
author_rank: Super Contributor
timestamp: July 25, 2023, 09:53:57 pm
content: Quote from: Scrts on July 25, 2023, 08:22:38 pmAs far as I can see from the schematic 7.5V is only used as impu by the two LDO units.I have to measure the absorbed current, I'd like to understand if there is some dead capacitor (aged esr?) or something that causes more current to be absorbed.3.3V is the CPU vcore, coldfire CPUs shouldn't eat so much current (1.5A is the LDO limit, anyway)There is a little heatsink on the 3.3V-LDO, the one that overheats, but I cannot believe the board was intentionally designed with an LDO that overheats to the point you cannot touch it with your fingers.I don't know if the board was trashed because damaged or simply because obsolete, it was dirty and full of dust.I cleaned it and seems correctly responding on the serial line, but I need to write and run a full diagnostic to be sure.

author: DavidAlfa
author_rank: Super Contributor
timestamp: July 25, 2023, 10:36:35 pm
content: Is it really overheating (Output voltage dropping or unstable) or just hot?Remember most parts are can sit at 100°C just fine.Given you found it in a dumpster, it wouldn't be a surprise that something was broken, consuming too much current?

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 25, 2023, 10:42:01 pm
content: Quote from: DiTBho on July 25, 2023, 09:53:57 pmThis was very common before switching regulators became ubiquitous.And an LDO dissipating a few W will get extremely hot. That's normal.Whether the heatsink was undersized or not for this board, I don't know. If you have a thermometer, you could measure the actual temperature. 70°C would already feel burning hot to the touch while perfectly fine for the regulator itself.You could replace both regulators with switching regulators if you want to increase efficiency dramatically (and decrease heat at the same time).

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 26, 2023, 03:29:28 am
content: Quote from: DiTBho on July 25, 2023, 09:53:57 pmDid you measure current IN and current OUT ?The fact it has a heatsink, confirms they knew they had a heat issue'Too hot to touch' is not actually that hot - it can be 55~65'C, which is fine for semiconductors.If we take a ballpark of 25'C/W for a modest heatsink, then just 400mA of load will give a 67'C heatsink temperature at nominal room temp.

author: mariush
author_rank: Super Contributor
timestamp: July 26, 2023, 08:30:17 am
content: Another solution would be to replace LT1086 (1.5A regulator) with LT1085 or LM1085 or *1085 (3A linear regulator) and then power the LT1086-3.3 with the output of the 5v regulator.The 1085 regulator will dissipate (7.5v - 5v ) x current where current is 3.3v current + 5.0v current so it will need a heatsink as it will produce more heat, (7.5v-5)x3 = 7.5w instead of 3.75w but now the 3.3v regulator will produce less heat because it's now dissipating (5v - 3.3v) x 1.5A = max 2.55 watts instead of (7.5-3.3)x1.5 = 6.3 watts.As you won't have 1.5A of current on 5v and 1.5A of current on 3.3v, the actual power dissipation on 5v regulator will be less.

author: DiTBho
author_rank: Super Contributor
timestamp: July 26, 2023, 09:02:11 am
content: I haven't yet measured the current_in (from the PSU).Anyway, this is a pic of the two LDO units with the heatsink.It doesn't seem well sized to meI will also measure the surface temperature, on the metal, with a thermometer.

author: DiTBho
author_rank: Super Contributor
timestamp: July 26, 2023, 09:27:49 am
content: Quote from: DavidAlfa on July 25, 2023, 10:36:35 pmThe user manual of the board tells you can use a PSU from 7.5V to 9VWith 9V, the 3.3V output voltage drops to 2.8V, the LDO smells of something that is going to make fire and the CPU becomes unstable.With 7.5V, the 3.3V output doesn't drop (3.1V is ok), the CPU seems ok, but the LDO is very hot.I cannot use the amperometry part of my multimeter, I am waiting for a new laboratory PSU, that will come with an integrated amperemeter, so I will know the current absorbed by the whole board.Meanwhile, I am fixing my multimeter. No big deal, but I have that kind of bad luck for which a tongue of the ammeter bushing broke: to the point that it "dances" and does not make good contact: it must be unsoldered, and replaced. I'll do it in the afternoon.

author: mariush
author_rank: Super Contributor
timestamp: July 26, 2023, 10:25:32 am
content: You could replace the heatsinks with bigger ones, seems like you have vertical space.some examples :https://www.digikey.com/en/products/detail/assmann-wsw-components/V2006B/3476155orhttps://www.digikey.com/en/products/detail/aavid-thermal-division-of-boyd-corporation/504102B00000G/5832You should measure the output current of the 3.3v regulator. If you don't have a multimeter that can measure current, just desolder the Vout pin of the meter and place a 0.1 ohm resistor in series, between the output voltage pin of the meter and the through hole on the board. Now you can measure the voltage drop across the 0.1 ohm resistor (10 1ohm resistors in parallel = 0.1 ohm or a few sub 1 ohm resistors) with the multimeter in voltage mode, 0.1v drop on the 0.1 ohm resistor would equal 1A of current.That heatsink definitely isn't good for 7.5v - 3.3v at high currents.Also :based on the picture and ETHERNET written above the 3.3v regulator, it makes me think that regulator would be for the DV9008F ethernet chip above, but that chip works on 5v and has a maximum power dissipation of 500mW.Could it be that someone accidentally switched the voltage regulators around? Check the output voltage trace, see where it goes, to what chips.The AM29LV004 flash chips are 2.7v - 3.6v devices, maybe the 3.3v regulator is for those....see if those get hot, maybe one of those is going bad or there's some shorted ceramic capacitor around them.Don't blindly switch the regulators as that could make 5v go on 3.3v output and damage chips but check this possible scenarioSee attached datasheet if you want, you see the chip runs on 5v (and i would expect it to work on 5v as it's a ISA plug and play ethernet controller and ISA had 5v, -5v and 12v, but no 3.3v)

author: srb1954
author_rank: Super Contributor
timestamp: July 26, 2023, 11:37:41 am
content: Quote from: DiTBho on July 26, 2023, 09:27:49 amIf the 3V3 drops when the input supply is 9V but not at 7.5V it suggests that the regulator is going thermal shut-down due to excessive power dissipation.Looking at the size of the heatsink I would estimate that something in excess of 750mA would need to be drawn from the 3V3 supply or the LT1086 to go into thermal shut-down . This current must be going into a faulty chip or faulty capacitor, which would get quite warm. Are there any other devices on the board that are getting excessively hot i.e. hotter than you can touch?At this stage I wouldn't change the input voltage or add additional heatsinking to the LT1086 until you determine what is drawing the excessive current. Dropping the input voltage or adding more heatsinking will allow the LT1086 to deliver even more current which could cause more circuit damage unless you use a current limited bench supply as the input source.

author: DiTBho
author_rank: Super Contributor
timestamp: July 28, 2023, 10:51:33 pm
content: My new power supply arrived today. It can measure the current to the load.With 7.5V, the total absorbed current is 550mA (the LDO 3.3V is hot)With 6.0V, the total absorbed current is 300mA (the LDO units are both cold)I haven't touched any hot component, except the LDO at 3.3VI've also ordered a thermocouple kit for my multimeter, it should arrive tomorrow

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 29, 2023, 12:10:25 am
content: Quote from: DiTBho on July 28, 2023, 10:51:33 pmWhat else is being powered from the supply ?That large change in current, is certainly abnormal for regulators, and is even worse than a resistive load.6/300m = 20 Ohms 7.5/550m = 13.636 Ohms.Measure the current into the 3v3 LDO to be sure, and the output voltage, (eg by cut of a trace) as it may be time to replace it.

author: srb1954
author_rank: Super Contributor
timestamp: July 29, 2023, 01:33:01 am
content: Quote from: DiTBho on July 28, 2023, 10:51:33 pmThis is a very odd result!To a first approximation the current drawn by the LDOs should remain constant as the input supply voltage varies. Providing that a constant supply voltage is being distributed to the circuitry powered by the regulators that circuitry would be expected to have a relatively constant current draw and that constant current is simply passed through a linear regulator effectively presenting a constant current load to the incoming supply.At this stage I would suspect that the 3.3V LDO itself is faulty. You could check that the O/P voltage is constant as the I/P voltage is varied. If it is possible lift the O/P leg of both of the LDOs so that there is no external load on them and measure the O/P voltage and quiescent current drain. The quiescent current for an unloaded regulator should be under 10mA.

author: DiTBho
author_rank: Super Contributor
timestamp: July 29, 2023, 09:28:43 am
content: Quote from: srb1954 on July 29, 2023, 01:33:01 amExactly my feelings: it doesn't make look good.I am going to unsolder the LDO-3.3V and test it on a test bench with simulated loads with armored resistors, values such as to simulate 200..500mA on 3.3V, and see its output and temperature.Meanwhile, I have already ordered a pair of new LDO-3.3V units.One shocking thing: in the box where I found the boards in the dumpster, there were only 12V power supplies!!!What if the previous owner had powered the board with 12V while the manual says "no more than 9V"?I don't know how these LDO drives age and I don't know how they break from overheating for a long time.

author: asmi
author_rank: Super Contributor
timestamp: July 29, 2023, 11:06:12 pm
content: This is why I avoid LDOs like a plague, and only use them in two cases - 1) when the current consumption is negligible (50 mA or less), or 2) when I need extra clean power lines to power sensitive analog circuits (like ADC, DAC, PLL, etc.) - and in the second case I tend to combine LDO with a buck pre-regulator to minimize heating and maximize efficiency.

author: DiTBho
author_rank: Super Contributor
timestamp: July 29, 2023, 11:24:22 pm
content: Quote from: asmi on July 29, 2023, 11:06:12 pmas far as I have seen, all the old Motorola and Arnewsh Coldfire v1 boards (1995-2004) are made with two LDO units in parallelIt doesn't make sense for me, except for the ultra low power m68k portable devices, which consumes less than 40mA. You see an LDO in their evailuation boards.What would you use to power supply a CPU core + DRAM + Ethernet core? say, ~200..300mA in total.I am really tempted to unsolder both the two LDO units, and to design an external circuit(1)Quote from: asmi on July 29, 2023, 11:06:12 pm(1) hybrid switching module followed by an LDO with low voltage drop (say ~ 1V or less), with the LDO used to "filter" the residual ripple.Do you mean something like this?

author: asmi
author_rank: Super Contributor
timestamp: July 30, 2023, 12:32:40 am
content: Quote from: DiTBho on July 29, 2023, 11:24:22 pmYou have to remember that back in a day switching converters were much more complex to design as they tended to require much more external components. Nowadays there are buck modules which only require a pair of external resistors to set the output voltage, and a couple of capacitors for filtering - which is not that much different from LDOs.Quote from: DiTBho on July 29, 2023, 11:24:22 pmI typically use buck converters for such tasks.Quote from: DiTBho on July 29, 2023, 11:24:22 pmThis kind of arrangement I typically use to power sensitive circuits like PLL or ADC/DAC. If I need a low voltage rail (1.5 V and below), I prefer using split-rail LDOs with a separate bias pin (for example TPS74901) - they have very low dropout voltage, which is important at lower voltage because the current tend to be higher. For example abovementioned TPS74901 has a dropout voltage of only 0.28 V MAX at 3 Amps of current, which means at 1 V output you can have just 1.3 V at the input and have efficiency of 3 W / 3.9 W ≈ 77%, which is super-high for LDOs (it's not bad even for switching DC-DC converters).But that is for sensitive analog stuff. For regular digital devices having just a buck is more than enough for most cases as more-to-less modern buck converters can have very low ripple voltage (<50 mV is fairly routine, but <10 mV is also very possible).

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 30, 2023, 02:24:45 am
content: I see no reason for this case not to use switching regulators as I suggested earlier.There are even drop-in replacement modules, to replace classic 3-pin linear regs with a buck converter.Here's an example design from TI:https://www.ti.com/tool/TIDA-01450I'm sure there are such ready-made modules out there.

author: DiTBho
author_rank: Super Contributor
timestamp: July 30, 2023, 10:26:50 am
content: Quote from: PCB.Wiz on July 29, 2023, 12:10:25 am+5.0V --> { DM9008F, MC145407DW, TL7705ACD, ispLSI2032V-100LJ, LED_GREEN, LED_RED, OSC_20MHZ }+3.3V --> { CPU_core_MCF5307, AM29LV004T-100EC_FLASH_HI, AM29LV004T-100EC_FLASH_LO, MC74LCX16245DT, MC74LCX244DW, MC74LCX244DW, MC74LCX16244DT, MCM69F737TQ11, KMM366S104BTN-GO_SDRAM }

author: DiTBho
author_rank: Super Contributor
timestamp: July 30, 2023, 10:29:04 am
content: Quote from: asmi on July 30, 2023, 12:32:40 amok, thanks!

author: DiTBho
author_rank: Super Contributor
timestamp: July 30, 2023, 10:35:58 am
content: is it possible and reasonable to measure the ESR of smd capacitors?There are a lot (~40) of smd capacitors, { ~25x0.1 uF, ~15x0.01 uF }.

author: DiTBho
author_rank: Super Contributor
timestamp: July 30, 2023, 10:38:02 am
content: plus, qty=6, 10 uF tantalium capacitors

author: SiliconWizard
author_rank: Super Contributor
timestamp: July 30, 2023, 08:44:41 pm
content: Quote from: DiTBho on July 30, 2023, 10:35:58 amIn-circuit? I don't know of any reliable way.But most of the caps from what I can tell are ceramic caps, there's very low probability their ESR would grow over time. They can fail altogether due to cracking, but that's not something I would normally fret over unless you noticed some very odd behavior that could be explained by bad decoupling.So you could just desolder the electrolytic cap and tantalum caps (there are just a few), test them and resolder them or replace them if not within specs.If you don't have a meter than can measure ESR reliably, I would just not bother and replace them directly with new ones, there are just a handful of these.And again, don't bother with the ceramic caps, at least as a first intention.