Instructables

Diode question

I need help with diode. Actually a question: if you put diodes in series for example 2 diodes of 300 V will they act as one diode of 600 V?

nf1196 years ago
The voltage rating on a diode is it's breakdown voltage (which could be said as the reverse biased voltage drop) - the voltage required for the diode to switch from an inductor to a conductor in a reverse biased configuration - that is the voltage require for current to flow in the reverse direction. So, in short, yes, if the 300 V you are talking about is the reverse biased breakdown voltage of those diodes - which is very likely since Power Zener Diodes have a 300 V breakdown voltage. What are you using the diode for? I ask because the voltage rating for diodes rarely matters in most applications.
gmoon nf1196 years ago
Partly true; it more-or-less doubles the reverse blocking voltage, but doesn't double the forward current capacity. Two diodes in parallel will do that.

Unless the diodes are well-matched, the characteristics in each case won't really be doubled. In cases where the diodes are really pushed to their ratings, then voltage sharing resistors need to be added.

The same approach is used when two capacitors are connected in series to increase the current capacity. Unless "sharing resistors" are added, one cap will always carry a higher burden, leading to premature failure.

The series diodes forward voltage drop would be cumulative (as you thought, guyfrom7up.)
fig2_6.gifseriescap.gif
I would like to know how these "voltage sharing resistors" are supposed to work, since going by your diagram if the voltage supply is AC then the resistors are providing a current path when the diodes are off, you would not get rectification.
Reference 1
Reference 2

I've never had cause to explore this with diodes, so I can't elaborate on their effects; but their use with caps is pretty common...
gmoon gmoon6 years ago
Nah, same thing with both references, they show a DC source. With AC the current would just go thru the resistors and not be blocked.
First, who said anything about AC? Or rectification? Not the OP. Not me...

AKAIK, either diagram only makes sense if it's showing a reverse bias. The resistance values needed to 'balance' two nearly identical diodes would be very small, and likely not effect rectification (place a 5M resistor across a rectifying diode, and see if if prevents operation.)

Neither reference qualifies their use to only DC, and I believe that interpreting the diagrams only (when no distinction is drawn in the text between DC and AC) is possibly a mistake...
When looking at the diagram, with reverse bias, ideally then the diodes are not really there, they're open. I'm left looking at two resistors, which will allow current flow in the reverse direction. In the Reference 2 example, that's 9.4M-ohms total, 4000V, only 0.43mA. Except, real world, the diodes are allowing limited current flow, 50mA is specified, so I guess talking about reverse current flow is moot.

You're using a voltage divider circuit to "share" between the two diodes. While that may increase effective reverse voltage ratings, some reality is involved. Between differences in the diodes, even the tolerances of the resistors, you will not be doubling the rating. I sure wouldn't stick two 300V diodes together like that and expect them to handle 598V. Maybe 350V, if really pushed 400V, but not up near 600V.

(BTW, why is it with the 2nd reference there is a "W" showing up where an ohm symbol should be?)

Oh, some form of current direction alteration is implied. Those diodes are going in a real circuit somewhere. If all that was desired was blocking of current in one direction always, with a current source always going in that direction, nothing works better than an open circuit.
I agree, I'm sure there must be some lost with the resistors, and reaching the theoretical limit is impossible (it always is) ... I see series diodes in amplifier power source rectifiers all the time (without the balancing resistors, since they aren't running at the limit.) They must be there for a reason. Whatever the increased voltage tolerance, it's significant, or they would have used a single, up-rated diode instead.
They must be there for a reason.
Diodes are cheap enough in bulk. If only one is used and it's bad (current goes straight thru) other components could get killed during a full-power test, two in series is insurance. Also reduces customer complaints, they may never know one went bad in use. Even a marginal ratings increase could mean they can use the same diode as elsewhere in the board, or a standard size in-plant, reducing inventory. Many reasons possible.
Diodes are cheap enough in bulk.

Well, we can at least agree on one point. ;-)

As for the rest, I still maintain that the reverse breakdown voltage is ~ double for two--just as the forward voltage drop for two diodes is ~ double. That's why the series diodes are added, to counter inductive and capacitative spiking.

So it's your turn to supply references...
One.
Another usage for rectifiers with avalanche capability is putting individual diodes in series to obtain higher reverse voltage capability. The voltage normally divides in proportion to the reverse resistance of each diode. Since there could be large variation in the reverse resistance, the voltage may not divide equally across the series string of diodes. A transient of sufficient amplitude will drive the voltage across one or more diodes into the breakdown region. A typical diode will exhibit catastrophic failure when this occurs. However, avalanche type diodes will cause the voltage to divide much more evenly. Also, the avalanche diode is capable of protecting itself by handling transient energy, providing it does not exceed the energy rating of the diode. Thus, whenever diodes are connected in series to increase the voltage rating, avalanche diodes should be used.

Don't expect double the rating from two in series.

Ah! You could be seeing 2 diodes in series to drop a voltage down, cheaper than using a secondary voltage regulator.

Heh, apparently it's very rare to use them as you say, at least from a design viewpoint. Can't find them like that at Wikipedia, nor anyplace else really. A search for "increase diode voltage rating" turns up this thread on the first page.

I'll try to search more, but right now there's a lightning storm blowing so...
A friend of my made a high voltage (in the tens of kVs) rectifier by putting cheap general purpose diodes in series. It worked well and none of the diodes failed.

I believe its is in a diode's operating definition that when the voltage rating is exceeded all that will do is allow the current to flow backward - this doesn't mean the diode is burn out - its just normal a operation of a diode. They will only burn out if we exceed the maximum current rating. Some people say high voltage = high amps based on ohm's law, however I doubt you are gonna be able to supply 6 million watts (600V 10 kA) - which is needed to burn out two general purpose diodes in series in a reverse biased configuration.
A friend of my made a high voltage (in the tens of kVs) rectifier by putting cheap general purpose diodes in series.
Why did he need one? You can find them in old microwave ovens. Was that really just a diode assembly, rectifying a high-voltage source, or part of a voltage multiplier circuit? Depending on the circuit, multiplier diodes only have to be rated for the incoming voltage or twice that. You can build a nifty high-voltage circuit with general-purpose diodes.

I believe its is in a diode's operating definition that when the voltage rating is exceeded all that will do is allow the current to flow backward - this doesn't mean the diode is burn out - its just normal a operation of a diode.
A Zener diode perhaps, but they're specially made for that, think there's a few others that can do that trick. But otherwise, at breakdown the diode will heat up, melt the little bit of silicon or whatever, and there goes the precise doped semiconductor regions, when it cools down what's left won't be a rectifier but will let current thru, both ways. With enough amperage the bit will vaporize, and no current will get thru.

Some people say high voltage = high amps based on ohm's law, however I doubt you are gonna be able to supply 6 million watts (600V 10 kA) - which is needed to burn out two general purpose diodes in series in a reverse biased configuration.
You have "general purpose" diodes that can handle 10,000 amps? Dang, the small general purpose ones I play with might only do a few amps. Where do you get those diodes?
For the 10 kA I used the listing from gmoon's reference 2. http://www.powerdesigners.com/InfoWeb/design_center/articles/Diodes/diodes.shtm

Two mechanisms can cause breakdown, namely avalanche multiplication and quantum mechanical tunneling of carriers through the bandgap. Neither of the two breakdown mechanisms is destructive. However heating caused by the large breakdown current and high breakdown voltage causes the diode to be destroyed unless sufficient heat sinking is provided.
http://ece-www.colorado.edu/~bart/book/book/chapter4/ch4_5.htm

So as I stated, most AC high voltage applications cannot supply a large current - which is needed, in conjunction with the high voltage, in order to heat up and destroy a diode.
For the 10 kA I used the listing from gmoon's reference 2.
Well so much for that "reference." A common 1n400x diode can only do 1A normally, resistive and inductive load, subtract 20% for capacitive load. 30A for a single stray short pulse. "High current" for a single diode is about 75A in an industrial-looking package. You can find "rectifiers," large units with many diodes, around 1000A, they're used for electroplating, large items or large batches I'd assume. Heck, I doubt you could even find a wire that can handle 10kA, for copper wire to handle only 300A requires a 4/0 (0000) stranded cable with 0.46" diameter, that's 0.166 square inches cross section, 10,000A would need 5.54 sq in, about 2.7" diameter! And that's just the copper, with insulation it'd be about 3 inches across.

Now, I don't know about you, but if I came across a single diode that needed a 3" copper cable to hook up to it, I sure wouldn't consider it "general purpose." Might not be one of the "special" types, but it sure wouldn't be used by a DIY-er for HV, let alone anything else.

You quoted:
Neither of the two breakdown mechanisms is destructive. However heating caused by the large breakdown current and high breakdown voltage causes the diode to be destroyed unless sufficient heat sinking is provided.
For the common 1N4001 the top temp of its operating range is 175o C, beyond that it goes real bad real fast. They're in an epoxy package, not known for shedding heat fast, most of the heat comes out on the leads. When they're talking about heat sinking, that refers to a diode made for a heatsink, like a TO-220, TO-247, or a stud diode. A regular epoxy-case axial-lead diode, the most common type used for homebrew, is not made for a heatsink. When they get current rammed thru them backwards they tend to die, quickly.

So as I stated, most AC high voltage applications cannot supply a large current - which is needed, in conjunction with the high voltage, in order to heat up and destroy a diode.
But... There are maximum power dissipation ratings for diodes, limits to how much power can go thru them. They're based on how much heating they can take before dying. The 1N400x series only does 3 watts. At a lowly 10,000 volts, they're maxed at only 0.3mA. Do you think 0.3 milli-amps is a large current? Because not much more, no matter which direction the current is going, will put them over their ratings and kill them.
Ok, it was bad judgment for me to trust the reference for the 10 kA. However, do we both agree that the breakdown is not the cause of the burn out? You need to be able to supply enough current also? If so, we need to know what Dantex is using them for in order to determine if it is gonna burn out because of breakdown or if it would even reach to the point of breakdown.
However, do we both agree that the breakdown is not the cause of the burn out? You need to be able to supply enough current also?
No. There are many types of diodes, among the regular and special types. While plain doped silicon is most often encountered, there are many constructions and chemistries, both historically and currently. For example, a micro-thin layer of oxide may be involved, and voltage, aka electrical potential, is a determining factor in oxide breakdown. This is seen with water, which is an oxide, when the voltage required for electrolysis is calculated. Once breakdown has occurred, with hardly any current at all, the oxide can be damaged. Likewise other chemistries can be altered. There may also be a thin place in a substrate where the actual breakdown will occur, a concentrated effect, which will cause overheating and damage at that spot. Semiconductors can be interesting and tricky, a MOSFET can handle large currents and be killed by a single stray spark.

In the ideal world frequently found in textbooks, you can make claims that proper heatsinking will keep a diode from dying. But out here in reality, if a device is not made for breadown, such as a Zener or avalanche diode, it is best to assume it is dead, for if it isn't obviously dead there's a good possibility it has suffered damage that will cause it to fail during use under load, no matter what current was involved.

...we need to know what Dantex is using them for in order to determine if it is gonna burn out because of breakdown or if it would even reach to the point of breakdown.
Nope. It's a simple question, will 2 diodes rated for 300V in series be good for 600V? In an ideal textbook world, yes. Out here in reality, given differences in diodes between manufacturers, even within the same batch by the same manufacturer, the blanket answer is no. And best of luck if you want to test a bunch to find two close enough that'll work.
So what is your answer yes or no?
To which question?
Dantex's question: if you put diodes in series for example 2 diodes of 300 V will they act as one diode of 600 V?
For an electrical exam question dealing with perfect parts in idealized circuits, the answer is yes. Reality says you won't know until you get up to 600V. Reality says don't risk it, buying more and better diodes is cheaper than replacing what gets trashed if it doesn't work.
They are pretty common, in some circles (a tiny sample):

http://schematicheaven.com/fenderamps/fender_bassman_ac568.pdf
http://schematicheaven.com/fenderamps/pa100.pdf
http://schematicheaven.com/fenderamps/fender_bandmaster_ad1269.pdf
http://schematicheaven.com/fenderamps/showman_6g14_schem.pdf

Don't expect double the rating from two in series.

Yeah, well theoretically, yes, as noted before. I'm still sticking with my original statement.

     There is now a sudden onset of current after the avalanche            breakdown voltage has been exceeded. Do not be confused into      thinking that this "breakdown" means that the diode has been      damaged. The process of avalanching itself is not      destructive....

From here. But certainly two series diodes can be damaged by too much current. Eventually. Given enough current. As can any single diode. Or 100 diodes in series...

Ah! You could be seeing 2 diodes in series to drop a voltage down, cheaper than using a secondary voltage regulator.

You're kidding, right? ;-) I'm sure we both know the difference between forward voltage drop and reverse breakdown. But the forward voltage drop is cumulative; I'm pretty sure the reverse breakdown voltage is, too...
Plasmana6 years ago
if you put diodes in series for example 2 diodes of 300 V will they act as one diode of 600 V?

Yes

guyfrom7up6 years ago
so if you were to have 2 zeners in series, lets say one's breakdown voltage is 20 and the other is 16 (random numbers here), would it be equiviilent to a zener with a breakdown voltage of 36?
Dantex (author) 6 years ago
I think that diode acts as a normal wire when current goes in right way and resistor in oposite
Basically right. The right way (forward bias) sends the current thru, within the amperage limit of the diode, while subtracting from the voltage what the diode takes to turn on, about 0.7V for an average one. The wrong way (reverse bias) it acts like a very high resistance, hardly letting anything thru at all. Too much reverse voltage leads to breakdown, the diode basically gives up, and it is ultra-rare for a normal diode to survive that.
nf1196 years ago
Before we continue with this "discussion" I think Dantex should verify that the voltage rating he is talking about is the reverse biased breakdown voltage (a part number would be nice). He also needs to provide the application of these diodes in order for us to see if it even matters and/or is the difference between theoretical calculation and real world situations will effect the performance and/or functionality of his circuit.
I don't think a specific senario (totally butchered that spelling) is necessary, it's a simple question of how can you increase the voltage handling of a diode. if you want part numbers, here's an example: if I were to put 2 1n4001 diodes in series, would it be able to handle 100 volts?
Yes the reverse biased breakdown voltage will be theoretically 100 volts. This means they should be able prevent current flow when a reverse voltage of 100 or less is applied.

However, this doesn't mean however they can't handle higher voltages. General purpose diodes have a froward voltage drop of .7 V, Does this mean they will burn out at higher voltages? No.

And the application does matter because, although the forward voltage drop is accurately stated, the true reverse biased breakdown voltage, which is like the reverse voltage drop is 20% higher than the reported rating. So if you are trying to build a rectifier, you will say under the rating, however in certain applications, the voltage does exceeds the ratings (ie. when the diodes are used in a high voltage trigger) then you'll need to consider the inaccuracy of the ratings and the fact that the true reverse biased breakdown voltage significantly varies from one diode to another.
gmoon nf1196 years ago
Clarification would be nice. Still, this "discussion" (your quotes) has been polite, repectful and on-topic. Hardly even qualifies as a spirited discussion, by instructables standards. Getting lots of emails, eh? Sorry--it has all been in a replies to your msg....
guyfrom7up6 years ago
I would like to know this answer, too. I'm think it just increases the voltage drop.
I'm think it just increases the voltage drop.
Agreed. If one put 600V through a 300V diode, it would burn out. There would still be 600V through each of the diodes, so they would still burn out.
forgesmith6 years ago
No. If we were talking about voltages in a circuit, then the voltage drops of components in series would add, if A used 300V and B used 300V then A & B in series would be fed 600V. But this is different. If we're not talking Zeners and related but regular diodes, then 300V is when it blows up. The diode acts like a voltage fuse. If you apply more than 300V the wrong way thru it dies. If there are two 300V diodes in series, the voltage drop across each one is minimal, each will see more than 300V trying to get thru. And blow.