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I have a 110W solar panel that I use to charge some LiFePO4 battery packs. The panel delivers 18-20V, so I have a DC buck converter between the panels and the batteries to bring the voltage down to 14.8V. To avoid having the batteries feed the buck converter when there is no feed from the solar panel, I put some diodes on the output side of the converter.

Initially I had two MUR460 diodes in parallel, but they got so hot that plastic case for the terminal block they were connected to melted.

According to the data sheet for the MUR460 diodes, they should be able to withstand a 4A average constant current, so two diodes was within the limits for the ~7A that a 110W panel should be able to deliver when the voltage is bucked down to 14.8V.

Due to the heat issue I replaced the two parallel diodes with three parallel, expecting that heat should no longer be an issue.

Those three parallel diodes still get very hot, not enough to melt the terminal block insulation, but still too hot to touch.

I will of course replace them with other diodes with a higher current rating, but I'm still curious: what am I missing here?

Are these diodes perhaps off-spec/bad, or is there something else that would explain why they get extremely hot? Did I just pick* the wrong kind of diodes, these are meant for high frequency switching, so are they perhaps not good for a constant forward DC current?

I would have thought that 3x diodes rated for 4A each should easily be able to carry a 6-7A forward current without producing a lot of heat.

*= the reason I picked those diodes was just that I happened to have a strip of them lying around in a drawer


Update: based on the advice in the answers and comments, I changed to a MBR3045CT schottky diode in TO220 packaging, and attached that to a heat sink. No more heat problems...

diode with heatsink

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    \$\begingroup\$ Well, you say that the diodes did withstand it... just not the plastic terminal block... so the datasheet wasn't lying. \$\endgroup\$ – user253751 Mar 22 at 11:07
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First of all, paralleling diode doesn't work very well. Unless the Vf are matched current will prefer one of the two. It sort of work, but not expect a 50% load sharing.

Fast diodes will work perfectly in DC, simply they are not optimized. Maybe they only have slightly more Vf (and are more expensive). Remember that dissipation in diodes is Vf times I. Schottky diodes also have substantial leakage which often predominates when you use them as switching diodes, but this is not the case.

110W at 18V (assuming the worst case) is slightly more than 6A; a cautious derating would choose an 8A silicon or Schottky rectifier for this application: this is a power oring condition (and I would strongly recommend a MOSFET and an oring controller for this amount of current) so let do some math on some plausible diode.

As a Schottky a choice would be for example an SBR1045, a chunky axial diode. Average I=6A, about Vf=0.4V at the operating point, about 25°C/W to ambient. That would dissipate 2.4W with a temperature rise of about 60°C. It could work, if you don't touch an 80°C diode body. Tj max in DC is 200°C so don't worry for the diode.

Now for the silicon choice: MUR820, the obvious stepup from your choice: the TO220 offers a 75°C/W free air dissipation, Vf at about 0.85V (at cold). It would dissipate 5.1W so there's no chance it would survive without heatsink (like most TO220 at their intended power).

In an oring controller situation, the main power loss would be the RDSon of the switching MOSFET. I'd take as a comparison a TO220 too but this would be in practice done with a DPAK if not with a smaller SOT package. But I'm lazy and I don't want to keep half a day choosing a MOSFET. Take the IRFZ24, I have a bag of these. Survives with 175°C at the junction. Nominal 70milliohm, about 2.5W of dissipation at 6A. Still not completely safe without heat sinking and in fact slightly worse than a Schottky. Well, it actually cost less than that diode, and it's not even a low gate drive MOSFET.

However these day steady state optimized MOSFET can go down to about 1 milliohm, so you'd be dissipating only a quarter of Watt.

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  • \$\begingroup\$ Thanks. Checking what was in stock at a local component supplier, I found this one: RHRP15120. I have some TO220 heatsinks, so I'll replace the 3x diode pack with one of those and a heatsink. Hopefully that should solve it. :) \$\endgroup\$ – KristoferA Mar 22 at 8:23
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    \$\begingroup\$ If I look at the datasheet, I(Vf) is a function of temperature. The diode that starts conducting the most will heat up more, which will increase its resistance, which in turn balances the currents between the two diodes. So there's no runaway effect, but you'd still end up with unequal temperatures. (In parallel applications, Vf will be matched, and the currents will differ) \$\endgroup\$ – MSalters Mar 22 at 16:34
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You need to do a heat transfer calculation. Even a few watts causes appreciable temperature rise if the heat cannot get out of the package quickly enough. What are you missing? A heatsink.

(The wikipedia article is long but in essence it is simple.) It goes like this:

  1. how many watts am I burning?
  2. what's the maximum junction temp of the device?
  3. what is the ambient temperature (max)?
  4. what's the thermal resistance, junction to free air, with no heatsink? (deg C/W)
  5. What size heatsink do I need to keep junction temperature acceptably low?

This is best done early in the design cycle as it always has mechanical implications (choice of package type, and how it is mounted.)

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From the diode datasheet, the voltage drop over the diode is 1.28V at 4A current. It will thus heat up at about 5 watts.

With lower currents the voltage drop is also less, so e.g 3A current has drop of 1.25V, so dissipation is less that 4 watts.

Also the voltage drop goes slightly down when the diode heats up, so the diode with lowest drop carries most current and when it heats up it will conduct even more current out of all the diodes in parallel.

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    \$\begingroup\$ Thanks, that's interesting. I guess that imbalance thing is the key thing, so I'll replace the pack of three with a larger diode + heatsink. \$\endgroup\$ – KristoferA Mar 22 at 8:24
  • \$\begingroup\$ That's probably the quickest and simplest solution. Another approach of course would be to switch on the output of the regulator using the input voltage from the panels, using a relay or a MOSFET. Then your power dissipation is close to zero. Fix the problem at source is better, when possible. \$\endgroup\$ – danmcb Mar 22 at 8:50
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You can only run parts at their rated power if they have really good heatsinks, see note 2. And at this power, the part will get really, really hot. If you aren't using a PWB with a few square inches of copper to dissipate the heat, solder the diode to a few inches of thick uninsulated bus wire (~12 AWG) on both sides. Keep the diode leads short.

Like this, except at a bigger scale, this is a tiny zener.

enter image description here

Or, use a bigger chassis mount diode with a proper heatsink.

https://www.onsemi.com/pdf/datasheet/mur420-d.pdf

Diodes in parallel is usually not the answer, as Justme explained.

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  • \$\begingroup\$ Thanks, I'll step up to a single large diode + heatsink. Hopefully that will work better. :) \$\endgroup\$ – KristoferA Mar 22 at 8:25
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To avoid having the batteries feed the buck converter when there is no feed from the solar panel, I put some diodes on the output side of the converter.

If you put the diode on the output, then the output voltage will drop by the forward voltage of the diode, which depends on current, so your charging voltage will not be accurate.

If you put the diode on the input, that problem is solved, but then the buck converter may still draw some power from the battery through its output and discharge it. So you have to check how much current it draws in this situation.

Higher voltage diodes tend to have higher voltage drop, so the RHRP15120 you got is even worse. So, try to get a 30-40V TO220 schottky diode instead. These or an equivalent part should be available from the local shop. It should drop less than 0.5V so a small heat sink will do the job.

The best solution is a low RdsON PMOS in the output, driven by a voltage comparator that turns it on if the input voltage of the buck is higher than the battery voltage. Make sure to add some hysteresis so the FET doesn't go linear.

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  • \$\begingroup\$ Thanks, I found a MBR3045 45V schottky diode in stock at one local supplier, so I'll try with one of those + heatsink. Unfortunately, the buck converter behaves a bit funny when powered through the output only, so I think a diode on the output side is a must. When backwards powered, it draws 300mA and the LED segment display on it starts flickering. I guess it is just not designed for that scenario... \$\endgroup\$ – KristoferA Mar 22 at 8:56
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Lots of good answers already, but one key point hasn't yet been emphasized: Rated power of diodes and resisters is often at 100 degrees C. And that will burn you if you touch it! So the fact that the components are very hot in use is not an indication that their rating is exceeded.

I do agree components will be more predicatable and have longer life if keep them cooler, but no need to keep them cool or near human body temperature.

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Okay here is my take on this.

  1. Diode(s) used possibly not fast enough.
  2. The package types on those diodes in my opinion terrible to heatsink.
  3. I`d experiment with something like a S30D30 salvaged from a broken ATX power supply.

Much easier to heatsink as well.

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  • \$\begingroup\$ Thanks. I am a bit confused by point #1: not fast enough. This is DC current that only switches twice per day: once at sunrise and once again at sunset. \$\endgroup\$ – KristoferA Mar 25 at 10:22
  • \$\begingroup\$ scrap comment point 1. I noticed you used a diode with adequate trr. It was probably just current rating and heatsink issue. \$\endgroup\$ – mnemonic Mar 26 at 12:01

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