# Is paralleling diodes a bad idea?

I'm trying to repair an 800W power supply (see my previous question on this.) One thing that gets me is that the design has two schottky diode packages (in TO-220) in parallel. I was always told this was A Bad Idea, but since they are thermally coupled to the same heatsink, does it present a problem in this instance? I've also noticed the same for the input bridge rectifier, two are used in parallel.

• Not an answer, but just a double check. Could it be that diodes are actually just a half bridge say with common cathode.
– user924
Nov 4, 2010 at 1:26
• Each package contains two diodes I think. Nov 4, 2010 at 7:56
• here's some useful video youtube.com/watch?v=WFHwjfhLk94 and youtube.com/watch?v=ZH4fs6xkWbk Oct 4, 2015 at 23:40

The issue with putting diodes in parallel is that as they heat up, their resistance decreases. As a result, that diode ends up taking on more current then the other diode, resulting in it heating up even more. As you can probably see, this cycle will cause thermal run away causing the diode to eventually burn if you give it enough current.

Now the fact that you couple them to the same heatsink will reduce this effect some, but I still would not recommend it. There are far too many unknowns that will affect this to not ever trust it, especially in a commercial product.

Now for the case of this power supply you are looking at, it may very well be that they spent the time to get the diodes matched as closely as possible and allow the heatsink to keep them at about the same temperature.

It may also be that they are running the diodes far under their capacity and they put the second one in parallel so that they aren't always running them near max capacity, but I find this unlikely.

• I have a professor who does this often, he will take 1000 diodes, literally, and measure their VI curve. Then he will strain them to have a very precise match. He sorts them into matching bins. Then he keeps them heat-sinking together. Nov 4, 2010 at 3:42
• @Kor: Don't diodes usually fail closed? True, it might "open" given enough power, but I'd think it would of taken something else out by then. Nov 4, 2010 at 18:04
• Good point, I forget that. You can get them that fail open, but sometimes we all say stupid things, I just do it more than most. Nov 4, 2010 at 18:08
• In most dual TO-220 packages, not only are the diodes thermally coupled together, they are generally both on the same die (e.g. made from the same chunk of silicon, in the same manufacturing run). This means they typically have very closely matching characteristics, which is why it works. Nov 5, 2010 at 5:46
• I took it as him saying it was 2 separate packages by him saying "two schottky diode packages". Nov 5, 2010 at 5:48

If you put a low value resistor, for example 1 ohm or 1/2 ohm, something like that, in series with each diode, and then parallel those assemblies, the resistors help to keep the load even between the two diodes. If one diode starts to take more of the load current (as it would with thermal run-away), the IR drop on the resistor lowers the voltage for that diode, tending to push the current back down.

The resistors need to be rated for whatever I^2*R loss they incur, and this usually means multi-watt ratings. Fortunately this sort of thing is typically only encountered in power supplies, where the inductance associated with wire-wound resistors isn't a bad thing. It's generally not a problem to find 0.1 ohm, 0.25 ohm, etc, in 5W, on up.

• Your answer is nice. Do need to make sure the resistor is enough to stabilize the system, but it is a very small resistance at high current, so easy to do. Nov 4, 2010 at 4:52
• So let's say I wanted to size for 40A current draw, so I'm going to put 4 x 10A diodes in parallel. Would I literally just calculate a 2W resistor value for each diode as: 2W/(40A/4 diodes) = 0.2 Ohms? Oct 3, 2017 at 21:46
• Also: quick follow-up: I see many large diodes in TO-220 packages--some rated for 100A or more. Can they really handle 100A in one of these packages or are heat sinks required? How do I know when heat sinks are required? Ex: here's a 40A diode, but I don't know if I have to use a heat sink or not and the datasheet doesn't mention it: st.com/content/ccc/resource/technical/document/datasheet/group3/… Oct 3, 2017 at 21:48
• Any diodes running more than 2-3A definitely need a heatsink. Just think about the power (in watts) that will be dissipated as heat; at 20A and a 1V forward voltage drop, the diode would be dissipating 20W as heat. The more watts dissipated, the larger the heatsink needed. There are calculations available for that. Jul 26, 2020 at 11:18

It's not ideal but in practice you can usually get away with it, especially if they are thermally coupled. If they aren't the potential problem is that silicon's -ve temperature coefficient could make one 'hog' more of the current, however in practice they will tend to both heat up at the same rate, and the slope resistance is never zero, so you will still get current sharing even when one is hotter.

Silicon Carbide (SiC) diodes can be paralleled without problem.

As one diode heat up its resistance increases so other diodes taking up more current. Current equalizing is guaranted by SIC semiconductor material. You don't have to worry about thermal runaway. However they are still expensive.

1) If what you have is a dual package because it was in stock or used used elsewhere on your BOM etc. then definitely go ahead and parallel. There is nothing to loose.

2) Some less common types of diodes like silicon carbide increase voltage with temperature and therefore can be reasonably well paralleled.

From the switching power world, I have seen paralleled diodes and bridges as you've described. The hope is that with matched heatsinking and small part-to-part variation, the devices will share load without any sort of external balancing. There are no guarantees, of course, so each of the devices has to be rated to handle the full load current else there can be "issues" (to put it mildly)...

One thing I don't see discussed is the peak to average current. The parallel rectifiers have a capacitor on the DC side that is charged up somewhat close to the peak voltage. Therefore the rectifiers conduct only when the peak AC voltage exceeds the capacitor's voltage plus 1 diode drop. If the average load on the DC side is say 1 amp, then the rectifiers are going to conduct peak currents several times that much.

So the rectifiers see high peak currents, and these make the low internal resistance of the diode junctions look to be equalizing the current between the paralleled rectifiers, just as someone else's suggestion to add external resistors to balance the current. I've seen this happen with LEDs.

So it seems to me that besides (or instead of) matching diodes for forward voltage, they should be matched for forward voltage at their peak current.

Here's a .PDF from ST with the nitty gritty. After I perused it I took away that if forward voltage difference is less than 40 mV it's ok to parallel diodes.

I was also thought that paralleling diodes is a bad idea, but thinking about it I'm not sure. If the current in diode 1 is greater than that in diode 2, the forward voltage drop (Vf) of the former will increase, which means more current will take the "shorter" route of diode 2. This looks like an auto-regulating circuit which keeps itself balanced.
Condition is that Vf-If characteristics should be comparable, which should be the case for diodes from the same batch.

edit
For Vf-If characteristics to be the same both devices should have the same temperature, so should be thermally coupled, best in the same package.

• It unfortunately doesn't work that way. If one diode heats up, the forward voltage falls, allowing it to conduct more current, heating it up more, causing the forward voltage to fall... eventually leading to diode failure. Dec 2, 2010 at 15:12
• @Thomas: That's why I mention thermal coupling. If temperatures are the same, the diode carrying the higher current will have the higher voltage drop. Dec 2, 2010 at 15:20
• Sorry, this answer seems woefully incomplete without also at least mentioning the transient thermal behaviour of the diodes. Sep 4, 2012 at 22:36
• If the diodes are parallel they have the same voltage drop. Jun 28, 2013 at 6:25

Plain and simple, two mismatched diodes do not share the current equally. There is no failure or runaway unless driven by a voltage source with unlimited current. A scenario popular with academics with no practical app. In reality a power supply may have a 5 Amp load and if really mismatched almost 5 Amps will flow thru one and nearly none in the other.

• If you want to parallel two diodes they must be thermally coupled or they must not operate close to the limit. If they are not you can easily destroy them. Current sharing depends strongly on their iv characteristics so it all depends on particular diodes. Jun 28, 2013 at 6:23

You can use series resistors to improve matching. R series would be calculated by R=0.6/Imax. Another method I have used with success is to find an appropriately rated bridge rectifier module and short the AC terminals together ("cathode") along with using the + terminal ("anode"). In the latter case, you don't need to use equalizing resistors for most applications since both bridge diodes are on the same die and share the same thermal envrironment. For demanding applications, you can still add equalizing resistors to the bridge rectifier solution by inserting them in series with each AC leg. Finally, smaller ohmic value series resistors will get the job done with the bridge rectifier module solution since the diodes are very similar to begin with. Hence, the bridge rectifier solution is more energy efficient than the 2 discrete diode solution at 60Hz power line frequency.

Crispy17

Everyone seems to be neglecting the small but still significant amount of inherent resistance of the paralleled [or dualled] Schottky diodes' leads. If one diode begins to 'hog' more current than its paralleled partner[s] then the increased resistance of its package's leads [having a positive temperature coefficient (unlike the junctions) n.b.] will automatically lessen the voltage driving against the junction[s]. This continual see-sawing [shared] action between the two or more diodes will average-out the current over time provided that the amount of current demand isn't excessive of course.

• For leads to counter thermal runaway effects, their voltage drop should be the same order of magnitude as the voltage drop on the junction. This may be the case for high currents and unusually long leads, but not in typical designs. Mar 16, 2017 at 16:18