Schottky diodes in parallel

[Diver300]

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.

I agree that is so stupid. 3 million pennies is £30,000 and you can't get an EMC test done for that.

At least with the Schottky diodes the only real consideration is that the no diodes are overloaded. Adding an additional diode may make the current distribution even less even, but it still helps in keeping the current lower in the hottest diode.

 

[MrAl]

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?

Hi,

You are certainly right for noticing that.
In this case though we are talking about the total 'dynamic' resistance. That's the value of the equivalent resistance at the specified operating point. In this case we simply divide the diode measured voltage drop by the diode forward current, which of course is just Vd/Id. That's the resistance I was talking about. The internal resistance is a different measurement which as you say, has to be based on some other parameters.

The reason I chose this resistance is because that gives us a starting point for selecting the required series resistance. This is a fairly important point because if we make the resistance too small it does not compensate for the voltage difference between two diodes in parallel (two diodes would be considered the theoretical proper). If we make the resistance too large, we not only waste power we also may end up reducing the voltage so much that the device it powers no longer works. I'll illustrate this in my next post.

Thanks for mentioning this because it could cause confusion without some more information.

 

[MrAl]

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.



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.

Hello there Nigel and thanks for the reply.

I mentioned the calculation of the series diode because it is really impossible to just throw any old resistor value in there because the effect could go from not enough to way too much. An example follows.

We have one diode D1 dropping 0.8v and the other D2 dropping 0.9v with D1 current of 2 amps and D2 current of 1 amp. The formula in this case is:
R=(V2-V1)/(I1-I2)
and plugging in the values we get:
R=0.1 Ohms
which matches one of your suggestions.
Now right off we can see that if you used a 0.22 Ohm resistor, we would be dissipating more power than need be, but that's not the worst of it.

Next we have D1 dropping 0.8v again and D2 dropping 0.9v, but this time D1 current is 2 amps and D2 current is 1.9 amps. Plugging these values in we get:
R=1 Ohms.
and that is what it takes to equalize the two currents.

It is possible that we might want to go through another iteration of these calculations but you can see the difference. If we chose 0.22 Ohms for the first case we would be dissipating too much power, and if we chose 0.22 Ohms for the second case we would not be compensating enough.

An extreme case would be I1=100 amps and I2=80 amps, V1=1.8v and V2=2v, then we get:
R=0.01 Ohms.
Big difference, but not only that, if we go with 0.22 Ohms for this example we drop around 40 volts, and even with 0.1 Ohms we drop around 20 volts, and for a 30 volt power supply that would not be very good.

I happened to specialize in power supplies back in the day so I might have a lot to say on this subject

 

[Nigel Goodwin]

Hello there Nigel and thanks for the reply.

I mentioned the calculation of the series diode because it is really impossible to just throw any old resistor value in there because the effect could go from not enough to way too much.

A nice scholarly post, but of no real interest to this thread which is one specific example, 3x3A diodes in a Vestel TV.

While it might be impossible to throw "any old value" in, "any old SENSIBLE value" will be perfectly fine - and the two examples I gave are both excellent choices, depending which you have. A 0.22 ohm will obviously waste a tiny amount more power, but nothing of any significance.

Your 2A and 1.9A example is pretty ludicrous, that's already more than balanced enough - no issue with the diodes failing. The point of adding the resistors is to prevent one of the diodes going short, not to produce some theoretical perfect balance.

I would also suggest that you're considerably overestimating the actual current in the Vestel TV - I doubt the total current is as high as 3A - if I ever get one to do again, I'll see if I can measure it?.

I seem to have a vague recollection, of running one with just a single diode - as a matter of interest, while I was waiting for a delivery of diodes from RS - it ran the best part of two days perfectly fine, until the diodes arrived and I fitted three new ones.

 

[danadak]

Hello there Nigel and thanks for the reply.

I mentioned the calculation of the series diode because it is really impossible to just throw any old resistor value in there because the effect could go from not enough to way too much. An example follows.

We have one diode D1 dropping 0.8v and the other D2 dropping 0.9v with D1 current of 2 amps and D2 current of 1 amp. The formula in this case is:
R=(V2-V1)/(I1-I2)
and plugging in the values we get:
R=0.1 Ohms
which matches one of your suggestions.
Now right off we can see that if you used a 0.22 Ohm resistor, we would be dissipating more power than need be, but that's not the worst of it.

Next we have D1 dropping 0.8v again and D2 dropping 0.9v, but this time D1 current is 2 amps and D2 current is 1.9 amps. Plugging these values in we get:
R=1 Ohms.
and that is what it takes to equalize the two currents.

It is possible that we might want to go through another iteration of these calculations but you can see the difference. If we chose 0.22 Ohms for the first case we would be dissipating too much power, and if we chose 0.22 Ohms for the second case we would not be compensating enough.

An extreme case would be I1=100 amps and I2=80 amps, V1=1.8v and V2=2v, then we get:
R=0.01 Ohms.
Big difference, but not only that, if we go with 0.22 Ohms for this example we drop around 40 volts, and even with 0.1 Ohms we drop around 20 volts, and for a 30 volt power supply that would not be very good.

I happened to specialize in power supplies back in the day so I might have a lot to say on this subject

I have been experimenting with simulation, dealing with a I-V response that is exponential leads
to exponential variations. And I am not yet convinced how good the spice models are. Or how
well matched the lot variation is. I worked as a PE on a PMOS line, we had, what would be deemed
huge today, variations in parameters across a wafer. Due to gas flow dynamics in the furnaces. The
wafer ET test sites on wafer of limited value. I built a system to track variations in Fortran on a mainframe,
basically what one would call today big data approach. Produced some meaningful results and change. I
remember the linear designers talking about same in linear process. Stuff they had to design around.
And as a test engineer later we allowed wide variation in parametrics, thank the profession for using
these approaches to design to tighter standards. In the case of mixed sig IC. But I wonder how much of
this was fed back to the discrete guys, often overlooked in many organizations......no glamor.....sadly

Regards, Dana.

 

[Diver300]

I have been experimenting with simulation, dealing with a I-V response that is exponential leads
to exponential variations. And I am not yet convinced how good the spice models are. Or how
well matched the lot variation is.

I am fairly sure that the issue depends on how the internal resistance of the diodes varies with the current, and how much the diodes heat up, and how much the forward voltage drop falls with temperature.

I also strongly suspect that it takes a very good simulator to model the change characteristics in response to temperature and current.

 

[MrAl]

I am fairly sure that the issue depends on how the internal resistance of the diodes varies with the current, and how much the diodes heat up, and how much the forward voltage drop falls with temperature.

I also strongly suspect that it takes a very good simulator to model the change characteristics in response to temperature and current.

Hi,

Yes we would have to model the internal heating as well as everything else. That's the main culprit.

I think this kind of problem may be outside the realm of a simulator that's why I highly recommend measurements.
We can model these actions, but we'd have to model at least one diode based on temperature, voltage, and current measurements, then apply that to the problem at hand. We might have to do a handful of diodes or more to get a good average behavior.

 

[danadak]

Actually I think the spice standard for a diode is OK, its just that the
vendor supplied models omit a number of the parameters.

https://www.acsu.buffalo.edu/~wie/applet/spice_pndiode/spice_diode_table.html

If one is so inclined :

SPICE modeling of a Diode from Datasheet - YouSpice

Modeling in SPICE& a diode is not a trivial work. Although the operation of the diode is quite simple, extract a model from datasheet takes some time.Every component has its own syntax defined in SPICE , in the case of the diode: .model ModelName D (par1=a par2=b………parn=x) where par1 par2 …...

youspice.com

The diodes I have looked at ~ 1/2 the parameters are there. Note the simulator does in fact
handle junction T. I have a case into simetrix to find out why my attempts dont seem to be making
sense.


Regards, Dana.

 

[MrAl]

Actually I think the spice standard for a diode is OK, its just that the
vendor supplied models omit a number of the parameters.

https://www.acsu.buffalo.edu/~wie/applet/spice_pndiode/spice_diode_table.html

If one is so inclined :

SPICE modeling of a Diode from Datasheet - YouSpice

Modeling in SPICE& a diode is not a trivial work. Although the operation of the diode is quite simple, extract a model from datasheet takes some time.Every component has its own syntax defined in SPICE , in the case of the diode: .model ModelName D (par1=a par2=b………parn=x) where par1 par2 …...

youspice.com

The diodes I have looked at ~ 1/2 the parameters are there. Note the simulator does in fact
handle junction T. I have a case into simetrix to find out why my attempts dont seem to be making
sense.


Regards, Dana.

It's interesting that you are looking at this.
I took another look and found that for two parallel diodes using just one resistor in series with the diode that draws the most current, the current can be equalized perfectly. The question remains what happens when the temperature changes. I have a feeling that they go unmatched again, to a significant degree.

 

[danadak]

Here, with 100 mV of offset, then 0, in one diode, 1 ohm in one leg :

Still waiting on sim guys to answer a question to get a more exact sim w/o
the artificial diode Vth I put in with fixed V source versus a model Vth change,
which shows no effect.


Regards, Dana.

 

[MrAl]

Here, with 100 mV of offset, then 0, in one diode, 1 ohm in one leg :

View attachment 142701


Still waiting on sim guys to answer a question to get a more exact sim w/o
the artificial diode Vth I put in with fixed V source versus a model Vth change,
which shows no effect.


Regards, Dana.

Hi,

Well I was thinking more like putting the resistor in series with the diode that originally draws more current. That would be D2 in this drawing.
1 Ohm probably too large though.

 

[danadak]

.1 ohm and 1 ohm

 

[MrAl]

.1 ohm and 1 ohm

View attachment 142705

Hi,

So what is your conclusion?

What about 2 resistors of different values?

 

[danadak]

Nothing earth shattering yet. Still waiting on sim vendor to reply
to model Vth effect not working as expected.

That being said a fixed .5V offset produces very large difference,
and only get balanced with larger R, say 1 ohm. Keeping in mind
currents all < 1 A


Regards,. Dana.

 

[MrAl]

Hello,

I was thinking that if we had a resistor in series with both diodes and one resistor was slightly larger it would compensate for the diode that drew the most current, and also push the total temp co positive instead of negative for both diode/resistor combinations. Might be interesting.