Diode In4007 burns due to excessive current?

[moonstreat]

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Hello, I have a 2 position switch, will my automatic and manual mode, I used a diode to only use manual and automatic one side of another. the problem is that my diode burning after a while, I think it is because no current will be that if I put two diodes in series solve my problem?

 

[MrAl]

Hello there,

Actually, no, but it *might* help if you put two in parallel, but even then it is very hard to predict how well the diodes will share the current.
As a better solution, try using a 1N54xx type instead, which has 3 times the current rating. It's bigger too so it will cool better.

 

[Reloadron]

The 1N4007 has an IF(AV) of 1 amp. How much is the load you are applying? No, placing another diode in series will not allow you more current as it would be a series circuit and the current through the first would be the same as the current through the second diode. The solution would be a diode that can handle a higher current. Again, what is the load?

<EDIT> Mr. Al was quick! </EDIT>

Ron

 

[moonstreat]

The corrent should be 5A

 

[MrAl]

Hello again,


Oh wow, well you need a much higher rated diode then. You could do a quick search on the web or maybe someone else here can suggest a low cost part number for you to purchase.

 

[colin55]

Just put two 1N54xx diodes in parallel. They are only 17 cents each and every shop has them.

 

[MrAl]

Hi colin,

As i said in a previous post, it isnt that easy to predict the way the two diodes would share the current.
Remember the voltage has a negative temperature coefficient right? That basically means it's a little feedback system in itself where one diode wants all the current to itself.

 

[Reloadron]

In a nutshell:

Question:

Can you put two diodes in parallel in order to obtain the double load?

Answer:

As a rule, NO!

The main reason is thermal runaway:
When you warm up a diode, its "conductance" at any particular voltage increases.
(The Shockley diode equation gives the exact equation).
Even with two identical diodes in parallel, if the temperature of one diode was even slightly hotter than the other diode, more of the current would go through the hotter one. The one with more current would heat up more rapidly, lowering its conductance even more, and after a short time the hotter diode is hogging most of the current.

The final effect is that a diode will carry almost all the current, while the other stays almost unused.

There are a variety of things that cause two diodes to *not* be identical, which only speeds up the thermal runaway.
*the unavoidable imbalance between the two diodes voltage drops.
*(with an AC signal): the diode with slightly faster turn-on time will absorb more turn-on loss
*(with an AC signal): the diode with slightly slower turn-off time will absorb more turn-off loss


There is an obvious exception: use identical diodes, with exactly the same electrical and themal chracteristics. Individually selected diodes with voltage drop matched to less that one mV, and carefully assembled for thermal matching, may carry the same current, so effectively double the rating.
It is a very unstable condition, like keeping a coin vertical on a table.
Another arrangement is to add current sharing resistors in series to each diode, at least 0.3/0.4 V of additional voltage drop is required, so is a very unefficient solution.
Considering high frequency signals (as in switching power supply) adds even more problems, so the answer is always: use just one diode, leave the task of paralleling only to very skilled people dealing with such high currents that no single diode in the market can withstand.

Conclusion would be find a diode rated for a higher current. Since the max current would be about 5 amps I would be looking for a diode rated for maybe 10 amps. There is no mention of the voltage but if the voltage is low maybe a diode from an automotive alternator would work.

The above quotes taken from the Wiki and can be found here.

Ron

 

[colin55]

If a 1amp diode (700mA diode) lasted a fair length of time, two 3 amp diodes (2,000mA diodes) are going to be absolutley adequate for the situation. A 10 amp diode is going overboard.

 

[geboy]

why can't we replace diode with some other high power component?

 

[Reloadron]

If a 1amp diode (700mA diode) lasted a fair length of time, two 3 amp diodes (2,000mA diodes) are going to be absolutley adequate for the situation. A 10 amp diode is going overboard.

Hi Colin and I may be off base with this but if we parallel diodes, even carefully matched diodes we know they each will not have an identical forward voltage drop right? The diode with the lower voltage drop will begin to conduct a little harder (more current) than the diode it is parallel with. As that happens it will likely get warmer and even conduct more. So if rather than 50/50 they started out at 60/40 as the diode conducting more would heat faster and eventually conduct even more to a point of maybe 70/30 or 80/20? I don't know but it has me curious?

As to 10 Amps I based it on the OP stating the load was 5 amps. Diodes are cheap and rather than look for a 5 amp diode and run it at maximum I figured I would find a 10 amp diode and run with it. That or use 1/4 of a full wave bridge rated at something higher than 5 amps.

Ron

 

[MrAl]

Hi there Ron,


Yes, that's the little feedback system i was talking about in action. The other interesting thing, as if that wasnt enough, is that as the one diode we've been talking about heats up it takes more of the load current and so the second diode ends up getting less current than it did when it started out with, which (ha ha) means it actually gets cooler! Thus, one diode gets hotter, the other diode gets cooler, not a nice situation by any means. Depending on the surface area of the diode body (and lead heat conduction) the one diode could burn up completely if the current goes over it's rating.

I'm going to look into this a bit further if i can and try to see if there is some sanity in all this by choosing the diode ratings carefully. If it were me though i think i would go with a single device that can easily handle the full load current. That's the way most power equipment is designed. Then again, we do like to experiment from time to time dont we

You know what else is interesting about two parallel diodes...if we were to cool one off with some cooling spray the hotter diode would conduct a lot of current and maybe after we stopped cooling that one diode it would still never be able to reach a state where both diodes were conducting 50 percent, even with perfectly matched diodes! That's because one diode would have already been heated, and i dont think there is any coming back from that state without turning off the system because the cooler diode will never be able to conduct enough current to heat up like the other one did. And that is with diodes with curves that are matched perfectly in every possible way. I'd like to model this and see what happens for sure.

 

[tvtech]

You guys ain't seen nothing yet...three 56V 400Mw Zeners wired in Parallel to pass as a normal 1W 56V Zener.....

And the real problem was not the original perfectly capable 1W 56V Zener going faulty (the set was designed that way and had been working for around 5 years)....it was the first dingbat that attempted the repair...the Power Supply was running too high...thus the original Zener had far too much heat to handle. And failed.

And so did the "Triple Bypass" the "Tech" tried....Three Zeners in parallel. I ask you. Think normal Diodes are bad...

I got the set on my bench. Replaced both 47MF in the Power Supply....And of course put a new 1W 56V Zener back where it belonged(s).

All is well. Set was repaired by me around Two years ago....and going stronger than ever. How do I know....he lives around the corner from me. I would of heard by now if something was wrong...And yes he is still alive.

And yes it is a Panasonic 54 cm Chassis.

Cheers

 

[crutschow]

I did a simulation with two 1N4007s in parallel to a 2A load. I changed the parameters of one to give about a 19% mismatch in current at 25C (.914A and 1.086A). Then I interactively readjusted the temperatures based upon the 50C/W junction-to-ambient thermal constant to 25C air. From this the relative current difference settled to .853A and 1.147A (about 35%) at junction temperatures of 63.5C and 76.5C respectively.

The is interesting since it indicates that the current hogging between two diodes (even when mismatched) may not be catastrophic. Of course this is only a simulation and must be viewed with reservation, but it is an encouraging result.

 

[tvtech]

Not to be nasty and all...theoretical evaluations of stuff do not always work in real life 100%.

Theory is hugely important, models are great.

Practical gut feeling + experience tells you everything.

Cheers

 

[crutschow]

Not to be nasty and all...theoretical evaluations of stuff do not always work in real life 100%.

Theory is hugely important, models are great.

Practical gut feeling + experience tells you everything.

I certainly agree simulations are not 100%, but my experience has been that they are usually quite good, if the models are adequate. Many times I've found errors or anomalies in my circuit designs from simulation, that I would have not discovered otherwise until after I had built the circuit.

Practical gut feeling + experience may tell you a lot, but not everything.

 

[cr0sh]

I understand the reasoning behind not paralleling multiple diodes, due to heat issues on the individual diodes; but what would happen if you mounted both diodes close together and bonded them together with thermal-conductive epoxy, and then blobbed that onto a heatsink? Would it work better? Worse? I guess the idea I am proposing (but I wouldn't try this in real life - this is just a thought experiment) would be to set up a feedback mechanism, such that one heating up would heat the other up, and they would balance out... Although now that I think about it a little more, you would probably end up in the same scenario, this time with both diodes failing...

 

[crutschow]

Mounting them on the same heatsink would certainly minimize the thermal runaway issue. The main differences in current would then be due to inherent unit-to-unit differences in the diode forward drops at a given temperature and current.

 

[MrAl]

Hi again,


Yes, mounting them on the same heatsink with thermal epoxy connecting the two would help to some degree because that would make the two diodes look like one single die thermally, so the only difference then would be the two bulk resistances and curve.

The following is assuming the diodes are thermally insulated from each other the way most diodes are used...

I wrote up a few equations and used them to try to make some sense of it, and i found out that it looks like when one diode starts out heated more than the other still no thermal runaway occurs, but rather the cooler diode heats up little by little and eventually the main difference is in the diode curve or the bulk resistance. If both diodes have the same exact specs, the two reach the same temperature. I think that's encouraging too.
One of the problems would be if one of the diodes got too much of the bulk of the current and that's probably where the real problem lies. If the diodes are over sized (relative to 50 percent rating) by say 150 percent it may work out ok in most cases. That would mean that for say a 4 amp diode two 3 amp diodes are used for example.

With even slightly different characteristics such as one diode changing by 2.2mv/degC and the other by 2.3mv/degC (internally of course) i was seeing about a 3 deg C difference in the final equilibrium temperatures. With one diode with 0.16 bulk resistance and the other 0.20 bulk i was seeing around 13 deg C difference, but the current ratios were as far off as 64 percent in one diode and 36 percent in the other. That would mean that if both diodes were rated for 75 percent of full load they would survive.
If one diode had it's curve off more than the other plus an unfavorable bulk resistance the situation became worse however, but interestingly, if one diode had it's curve off in one direction and the other diode had it's resistance off in a favorable direction, the currents could be evened out again!
A second way the currents would be evened out again turned out to be if the one diode had it's curve off and it's bulk resistance off in a favorable direction.
The bottom line here is that depending on any of the possible combinations of curve and bulk resistance between two diodes some combinations will work much better than others, and of course some may fail.

So it looks like it's a bit of pot luck, if we get two diodes that work well together we may get nearly even current sharing, but the other extreme is we get two that work very unfavorably together and we get serious current differences.

The other interesting thing this brings out is if we have two diodes that work well together and then try to use one of them with a third unknown diode the new set may fail even though the first set worked fine. This also means that the two diodes dont necessarily have to be 'exactly the same', but they simply have to draw the same currents when in parallel.

Can you say, "Current Probe"

 

[moonstreat]

I put a diode with 3A and solved my problem so far ..