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Schottky diodes in parallel

Diver300

Well-Known Member
Most Helpful Member
I've just fixed, for the second time, a small TV with integrated DVD player. The power supply has three Schottky diodes in parallel as the main rectifier on the low voltage output of the SMPS. The first time I fixed it, I hadn't realised exactly what was wrong with the power supply and I changed the whole board, which cost about £20. It's one of these:- https://i.ebayimg.com/images/g/IS0AAOSwCi5kuYN1/s-l500.jpg
The three diodes are to the left of the transformer. They are the three with the cathodes near the axial inductor and the four electrolytic capacitors on the left of the board.

I later realised what had actually gone wrong was that the middle diode had failed short circuit.

This time, I found that the same diode on the new board had failed. When I swapped it for one of the working ones from the old board, it worked.

Ok, so there is a marginal design that can have a thermal runaway because Toshiba shouldn't have paralleled Schottky diodes.

The board is designed for various configurations. There is even space for a completely different transformer set at 45°, and its outline can be seen poking out from under the transformer that is actually fitted.

Between the failed diode and the one above it, there is space for a fourth diode in parallel. I've fitted a fourth diode as I had one from the old board.

Is that going to help or make things less reliable?
 
Paralleled diodes are not a good idea without some form of load sharing arrangement.

Basically one will carry more of the current than the other's and as you have found, fail.

Better off putting a single device with a larger current handling capacity.
 
Paralleled diodes are not a good idea without some form of load sharing arrangement.

Basically one will carry more of the current than the other's and as you have found, fail.

Better off putting a single device with a larger current handling capacity.
I understand those points. I would not have designed the circuit like that.

My question is about whether having four diodes will be more reliable than three. The change to four diodes is trivial as there is already space on the board. Other changes would be a lot more work.
 
My question is about whether having four diodes will be more reliable than three.
Probably not.
One diode will still tend to hog the current and short.

What's the current capacity of the diodes?
It would be better, if you could, to substitute one large diode with a rating equal to the three smaller diodes.

Alternately adding a small resistor (e.g. 0.1 ohm) in series with each diode would help equalize the current.
 
My question is about whether having four diodes will be more reliable than three.
You would be better off with four (or three) from the same batch, as there is likely to be less variation in characteristics than with ones from different batches.

If possible I'd also leave the leads long and space them above the PCB, splayed out for better airflow.

The centre one failing each time with the original three could point to more heat build up than with the outer ones, resulting in the failure?
 
I've just fixed, for the second time, a small TV with integrated DVD player. The power supply has three Schottky diodes in parallel as the main rectifier on the low voltage output of the SMPS. The first time I fixed it, I hadn't realised exactly what was wrong with the power supply and I changed the whole board, which cost about £20. It's one of these:- https://i.ebayimg.com/images/g/IS0AAOSwCi5kuYN1/s-l500.jpg
The three diodes are to the left of the transformer. They are the three with the cathodes near the axial inductor and the four electrolytic capacitors on the left of the board.

I later realised what had actually gone wrong was that the middle diode had failed short circuit.

This time, I found that the same diode on the new board had failed. When I swapped it for one of the working ones from the old board, it worked.

Ok, so there is a marginal design that can have a thermal runaway because Toshiba shouldn't have paralleled Schottky diodes.

The board is designed for various configurations. There is even space for a completely different transformer set at 45°, and its outline can be seen poking out from under the transformer that is actually fitted.

Between the failed diode and the one above it, there is space for a fourth diode in parallel. I've fitted a fourth diode as I had one from the old board.

Is that going to help or make things less reliable?
It's NOT a Toshiba, it's a Vestel - it's a design flaw, and a VERY, VERY common fault.

The flaw is putting rectifiers in parallel with no load sharing resistors, so they WILL fail, and they do.

There's no point fitting a fourth one, that's for a model that needs more power.

You should replace all three diodes, with ones out of the same packet (as they are likely to be closer matched), you can get the exact correct diodes from RS. You could even add series resistors while you're doing it, if you wanted - but I've never had them fail again after replacing them with diodes from RS out of the same packet (and I've repaired a LOT of these cheap Vestel sets.).
 
I've just fixed, for the second time, a small TV with integrated DVD player. The power supply has three Schottky diodes in parallel as the main rectifier on the low voltage output of the SMPS. The first time I fixed it, I hadn't realised exactly what was wrong with the power supply and I changed the whole board, which cost about £20. It's one of these:- https://i.ebayimg.com/images/g/IS0AAOSwCi5kuYN1/s-l500.jpg
The three diodes are to the left of the transformer. They are the three with the cathodes near the axial inductor and the four electrolytic capacitors on the left of the board.

I later realised what had actually gone wrong was that the middle diode had failed short circuit.

This time, I found that the same diode on the new board had failed. When I swapped it for one of the working ones from the old board, it worked.

Ok, so there is a marginal design that can have a thermal runaway because Toshiba shouldn't have paralleled Schottky diodes.

The board is designed for various configurations. There is even space for a completely different transformer set at 45°, and its outline can be seen poking out from under the transformer that is actually fitted.

Between the failed diode and the one above it, there is space for a fourth diode in parallel. I've fitted a fourth diode as I had one from the old board.

Is that going to help or make things less reliable?

Hi,

Geeze it looks like almost everything has been said already, shows how the collective intelligence of the group works so well.

I like the current sharing resistors idea put forth by Cruts, and the replacing all of them with diodes of the same batch by Nigel, and i will add just a little to that here.

First, the current sharing resistors is probably the best idea, but i also realize it's not always easy to do that due to limited space. If you can do that though, probably should do that.

If you can't, then default to the 'new' diodes idea all from the same batch, and there's one more little step depending on what measurement equipment you have.
What you can do is buy 20 diodes (not that expensive usually) and test each one of them for voltage vs current, maybe for three levels of currents, but mostly for a higher current level maybe 75 percent of the rated current.
Check that the voltage is nearly the same for each of the three or four diodes. That will ensure you get at least some matching.
Next, connect the winners from that test in parallel, that would be 3 or 4 diodes, or just start with 3 or even 2.
With 3 in parallel, you can run them at the expected current (say 75 percent of the combined ratings) and check to see if any one of them gets hotter than the others. It's hard to check the current, but if you have low value resistors like 0.02 Ohms or something like that you can take a look at the current for all three also.
If you find that they get fairly warm, add that fourth pre-tested diode and see if they remain cooler after that.
A good life test is to let them run overnight, say 12 hours, to see if they survive.

The whole problem, as you probably know, is that diodes can experience thermal runaway, and because the characteristic voltage goes down when one gets hot, the total impedance of that diode goes lower and thus more current flows though that diode. That in turn heats it up a bit more, which then causes that impedance to drop even lower, etc., etc., until the diode blows out.
We are lucky that there is a limit to this, but depending on the exact diode specs it could be too late and the diode blows out.

With parallel LED's we face the same problem. I had a flashlight that was made by a decent company but in this model they wired all the LED's in parallel. I think there were 10 of them. After the LED's warm up, one starts to draw more current, and because the heating is a somewhat gradual process, the LED current rises somewhat slowly, which in turn makes it get hotter slowly. This slow increase in current does not immediately blow out the LED, it makes it blink on and off, as amazing as that sounds. Eventually though, it just goes out.

Note that this usually happens when we run the diodes or LEDs near their full rating when they are in parallel. So two 1 amp diodes should not be run at 1.8 amps for example. For a rule of thumb, probably a 1.5 to 2 current ratio is good enough, but you can go lower if you like. So two diodes rated for 1 amp each should only be run at a max of 1.5 amps.
If I get a chance I will go over the math behind this.

Oh I almost forgot the single diode solution. That's a good one too, although the multiple diode solution spreads the heat out a little better.

Good luck with it.
 
Last edited:
It's NOT a Toshiba, it's a Vestel - it's a design flaw, and a VERY, VERY common fault.

The flaw is putting rectifiers in parallel with no load sharing resistors, so they WILL fail, and they do.

There's no point fitting a fourth one, that's for a model that needs more power.

You should replace all three diodes, with ones out of the same packet (as they are likely to be closer matched), you can get the exact correct diodes from RS. You could even add series resistors while you're doing it, if you wanted - but I've never had them fail again after replacing them with diodes from RS out of the same packet (and I've repaired a LOT of these cheap Vestel sets.).
I would assume that Vestel used diodes from the same batch anyhow, so why would replacements, all from one batch from RS, be any better? Is it down to component quality?
 
I would assume that Vestel used diodes from the same batch anyhow, so why would replacements, all from one batch from RS, be any better? Is it down to component quality?

It could be that they used a mix of batches, caused by a particular purchase being half of one batch and half of another. The best bet is to test a number of diodes from what you think is the same batch, and match them yourself. Follow that with a complete test.
 
I would assume that Vestel used diodes from the same batch anyhow, so why would replacements, all from one batch from RS, be any better? Is it down to component quality?
We've no idea what they used, or what quality of diodes they bought (other than would be the cheapest they could possibly find) - using good quality diodes from RS, out of the same packet, seems to work with no further issues. What you shouldn't do, is change just one.

I have repaired a couple of similar sets where I work now (for 'friends'), and replaced all three diodes with MBR340G's, because I have them in stock, and are a similar type of device.

But if anyone is concerned, you could always add a resistor in series with each diode, something like a 0.1 or 0.22 ohm should be enough?.
 
Thanks to everyone for the replies.

The consensus is that the best solution is to add small resistors in series with each diode, which will improve the situation.

The reason that resistors improve the situation is that if one diode has a lower forward voltage, because of manufacturing tolerance or temperature, the resistors will reduce, but not eliminate, the difference in current. The diode with the lower forward voltage will still take more current than the others, and dissipate more heat than the others, but not as much as it would have done without the resistors.

It's then a question of which goes up faster with temperature, the head generated in the diode of the heat lost through heat sinking. If a small rise in temperature causes 1 W extra electrical energy to be converted to heat in the diode, but the heatsink only takes away 0.9 W of power for that temperature rise, the diode will continue to get hotter and thermal runaway has started.

If the small rise in temperature causes 1 W of extra heat generated in the diode, but that same temperature rise causes 1.1 W of power to be removed by the heatsink, the system is thermally stable.

Adding resistors is one way to improve the stability, because they will reduce the extra power that the resistors generate, without affecting the heat that goes to the heatsink, so suitable resistors can easily swing the system from unstable to stable.

This leads me to the conclusion that adding extra diodes helps, but I don't know how much by. More diodes in parallel means that there is less resistance in parallel with the hottest diode. This lower resistance means that if the hottest diode decreases its forward voltage by a small amount, the change in current will be less than if there were fewer diodes, so the stability is improved.

I don't know how big this effect, and what series resistors would be "equivalent" to an additional diode in parallel. I think that it is very difficult to work out, and will probably depend on many factors.
 
Thanks to everyone for the replies.

The consensus is that the best solution is to add small resistors in series with each diode, which will improve the situation.

The reason that resistors improve the situation is that if one diode has a lower forward voltage, because of manufacturing tolerance or temperature, the resistors will reduce, but not eliminate, the difference in current. The diode with the lower forward voltage will still take more current than the others, and dissipate more heat than the others, but not as much as it would have done without the resistors.

It's then a question of which goes up faster with temperature, the head generated in the diode of the heat lost through heat sinking. If a small rise in temperature causes 1 W extra electrical energy to be converted to heat in the diode, but the heatsink only takes away 0.9 W of power for that temperature rise, the diode will continue to get hotter and thermal runaway has started.

If the small rise in temperature causes 1 W of extra heat generated in the diode, but that same temperature rise causes 1.1 W of power to be removed by the heatsink, the system is thermally stable.

Adding resistors is one way to improve the stability, because they will reduce the extra power that the resistors generate, without affecting the heat that goes to the heatsink, so suitable resistors can easily swing the system from unstable to stable.

This leads me to the conclusion that adding extra diodes helps, but I don't know how much by. More diodes in parallel means that there is less resistance in parallel with the hottest diode. This lower resistance means that if the hottest diode decreases its forward voltage by a small amount, the change in current will be less than if there were fewer diodes, so the stability is improved.

I don't know how big this effect, and what series resistors would be "equivalent" to an additional diode in parallel. I think that it is very difficult to work out, and will probably depend on many factors.

Hi again,

This reminds me of a quote, although I do not believe it is complete it still helps with this kind of situation.
The quote is:
"A single measurement is worth a thousand expert opinions".

In other words, measure the dang difference. If you place a series resistor with each diode you can easily measure the current in each diode.

In reality, you can not just stick any old value resistors in series, you have to calculate the value. You can do this by considering the impedance of the diodes at their expected operating current.
An example is if the diode drops 0.5v and is carrying 1 amp, then the impedance is 0.5 Ohms, and 20 percent of that is 0.1 Ohms. On the other hand, if the diode drops 0.5v and is carrying 10 amps, then the impedance is 10 times less than that and 20 percent of that is just 0.010 Ohms. This means that soldering, lead length, and circuit trace length and thickness could play a part in the distribution of current through each diode. Diodes connected with straight circuit traces with 90 degree corners would not be considered symmetrical, some of the traces would have to be changed so that each trace that connects to the one common point are is the same length and width.

There has been success in modeling the thermal vs voltage characteristics using a third degree algebraic equation, and that leads to more understanding of what goes on. In general though, if you do not intend to match the diodes yourself then you may have to use a significant resistance to balance all the diodes. This causes power loss of course.
The ratio of resistor resistance to diode impedance may have to be higher than 20 percent.

If you place small value resistors in series with each diode you should be able to measure the currents in each diode given either a test current or in the actual application if that is possible.
In some cases the transient response also has to be considered because if that causes one diode to heat up first, that diode becomes more vulnerable that way too.
 
Hello again,

I forgot to mention a couple other things.

First, Schottky diodes rated for 60v or under match in parallel better than those over that voltage, because thermal characteristics are better for those lower voltage diodes.

Second, if you can use SiC diodes instead of Si diodes you can get better matching because of the way the voltage drop inverts at higher currents. Instead of the common approximation of -2mv/degC it becomes less negative, which means the higher current diode eventually starts to draw a comparatively lower current. That means that the other diodes start to pick up more of the load current than they would have with regular Si diodes. That of course means better load sharing.
You'd have to look into whether or not you can use these types of diodes in that application.
 
In reality, you can not just stick any old value resistors in series, you have to calculate the value.

Rubbish - no need to calculate anything - simply a little common sense and choose a suitable low value resistor. As I mentioned above, 0.1 or 0.22 ohm would be fine.

If you were trying to current share with different types of diode, it's a bit more critical, but not at all for sharing with identical diodes.

But as I've also said, just fit three identical diodes from the same packet from RS, and I've never seen then fail again. If I was designing it, I'd include series resistors, but I've not seen any reason to add them as long as you use decent diodes.

This leads me to the conclusion that adding extra diodes helps, but I don't know how much by. More diodes in parallel means that there is less resistance in parallel with the hottest diode. This lower resistance means that if the hottest diode decreases its forward voltage by a small amount, the change in current will be less than if there were fewer diodes, so the stability is improved.

I would disagree, the lowest Vf one passes the most current, with the highest Vf one passing little or none.

Adding more diodes in parallel will make pretty well no difference at all, the lowest Vf one will still take most of the load.

Just good quality (reasonably) matched diodes, or add series resistors.

I'm also pretty dubious about the requirement for three diodes in the first place, it'd be interesting to see how two (or even just one) copes.
 
I would disagree, the lowest Vf one passes the most current, with the highest Vf one passing little or none.
It's a matter of degree. If you look at the graphs on page 3 of this:- https://www.diodes.com/assets/Datasheets/ds30135.pdf
you can see that a diode at 100 °C will pass around 2A at 0.5 V, while one at 20 °C will pass around 0.7 A at 0.5 V. That is a big difference, but 0.7 A is a big chunk out of 2 A and would reduce the stress on the hot one by quite a lot.
 
When I worked for an automotive-module manufacturing company, we were approached by one of the Big-3 for some cost reduction ideas, for their highest volume platform.
The idea they had dreamed of was to feed groups of 5 LEDs in parallel with a single resistor, instead of individual resistors, feeding from the battery voltage.
These LEDs were used to illuminate the instrument cluster, and as such they had a quite tight brightness-spread specification, which had to be met from minimum to full brightness. We were already receiving the LEDs binned for brightness at a particular current level.
When I objected that this bright idea, -pun intended- would add another uncontrolled variable to an already difficult spec, which for the amount of “cost savings” wasn’t worthwhile, I was sternly advised that we assembled over 3 million assemblies per year, and 3 million pennies still are 3 million pennies. Was I a team player or not?

Long story short: the idea got implemented. Within days the uneven brightness issues surfaced. Not going to describe the whole ordeal, but suffice to say that thanks to the ISO9001 and IATF 16949, we could not go back to the original configuration.

We had to pay the LED manufacturer to re-bin them according to forward voltage in addition to the brightness. Which caused the LED price to jump 2X. So much for the cost savings.

Yes, large companies can be THAT STUPID.
 
An example is if the diode drops 0.5v and is carrying 1 amp, then the impedance is 0.5 Ohms, and 20 percent of that is 0.1 Ohms. On the other hand, if the diode drops 0.5v and is carrying 10 amps, then the impedance is 10 times less than that and 20 percent of that is just 0.010 Ohms.

The junction drop is not really part of the forward resistance?

I'd measure the difference in voltage drop between eg. 0.1A and 0.2A (or appropriate currents for the diode operating range) to calculate the internal resistance?
 
The junction drop is not really part of the forward resistance?

I'd measure the difference in voltage drop between eg. 0.1A and 0.2A (or appropriate currents for the diode operating range) to calculate the internal resistance?
The internal resistance changes with current. It gets less as the current increases, which is one of the causes of runaway.
 
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