Current through diode.

[The Electrician]

These are features that have NEVER been pointed out before.

Otto Schade described this behavior in 1943, and he wasn't the first, but his paper is the classic paper referred to in most app notes from the rectifier manufacturers, such as the Motorola Rectifier Handbook. See the first .pdf attachment.

More recently, in 1974, Peter Richman gave another exposition of this behavior. See the second and third attachments.

And, of course, the chart on the second page of the GE spec sheet I posted, titled "Derating for shortened conduction angle" is derived from the fact that the diode current when feeding a capacitor input filter is typically rather narrow pulses. So, apparently they knew about this.

 

[colin55]

Yes, I take your point but I am referring to the fact that it has never been pointed out in a simple, clear and consice form for technicians to understand.
I have taught many groups and no text book has really made this clear.
No-one mentioned this fact in this discussion and no technican in any of our courses has realised this situation, before being told.
It's something to remember.

 

[The Electrician]

Yes, I take your point but I am referring to the fact that it has never been pointed out in a simple, clear and consice form for technicians to understand.
I have taught many groups and no text book has really made this clear.

You made an absolute statement, "These are features that have NEVER been pointed out before.", which is not true.

I don't know what textbooks you've been using, but every textbook I've seen that shows the somewhat triangular ripple voltage out of a capacitor input filter, also shows that the diode current is narrow pulses, and conduction only occurs near the peak of the sine wave of voltage.

Furthermore, the fact of peaky current pulses in capacitor input rectifiers is all over the web, for example: Electronic Transformers - Rectifiers with Capacitor-Input Filters

And, every manufacturer of rectifier diodes, such as Motorola, GE, General Instrument, etc., all have application notes that explain this in great detail. Motorola has an entire book, the Motorola Silicon Rectifier Manual, devoted to rectifier circuit behavior.

The information is widespread and readily available to anybody who cares to look for it.

I think your perception of the situation is more pessimistic than it really is.

No-one mentioned this fact in this discussion...

I mentioned in three posts in this thread that the diode current is more peaked than a pure half sine, and I even posted a scope capture showing that fact. I posted a GE spec sheet that shows how to determine the derating factor for the diode due to the fact that the current is a narrow pulse, rather than a full half sine.

...no technican in any of our courses has realised this situation, before being told.

Maybe they're all using the same bad textbook. Maybe they haven't been working for companies that build power related products.

Technicians don't usually design power supplies, so they probably didn't need to know it, and if they were ever taught it, it didn't sink in.

The production line techs at the company where I work don't need to know it, but our engineering techs do know it.

 

[The Electrician]

As I said before, it really doesn't matter if the supply is capacitor fed, from a transformer or delivers to a capacitive load or resistive load,
The end result is heat is generated.
In fact the diode conducts for only a very small portion of the cycle and during this time the current flow is very high.

It does matter. If the load is a resistive load (that means no capacitor across the load), then the diode does not conduct "...for only a very small portion of the cycle...". It conducts for the entire half cycle.

Certainly, heat is generated, but if the conduction angle is 180 degrees (conduction for the entire half cycle, not just a narrow pulse) as it will be with a pure resistive load, then the heating of the diode is substantially less.

This is the condition under which the manufacturer specifies a rating of 1 amp average.

 

[colin55]

I think you have read my comment incorrectly.
I said it does not matter if the load is capacitive or resistive, the end result is heat is generated. This is true.

Furthermore, if the load is resistive or capacitive, and the same wattage is being drawn from the supply, the heating of the diode will be slightly more in the case of a capacitive load due to the I squared R losses.

 

[The Electrician]

I think you have read my comment incorrectly.
I said it does not matter if the load is capacitive or resistive, the end result is heat is generated. This is true.

And I didn't disagree with you on this point; I said "Certainly, heat is generated...".

But you also said "..In fact the diode conducts for only a very small portion of the cycle and during this time the current flow is very high....", right after the previous statement without saying that you only intended this to apply to capacitive loads, which to me suggests that you apparently intended it to apply to both cases, capacitive load and resistive load; if you didn't, you should have said so. It isn't true for a resistive load.

Furthermore, if the load is resistive or capacitive, and the same wattage is being drawn from the supply, the heating of the diode will be slightly more in the case of a capacitive load due to the I squared R losses.

Using equal wattage drawn from the supply for comparison adds a complication because the current supplied to a resistive load is pulsating; a series of half sine waves, not smooth DC. Previously we had been discussing just the average current through the diode.

Also, the heating in the diode is more complicated to calculate than just the I^2*R losses. When the current in a diode is large, the voltage drop vs. current is no longer an exponential relationship. The intrinsic resistance of the silicon chip begins to dominate. For lower currents, the drop across the diode is nearly constant at the well-known .7 volts.

When the current in the diode is large, the resistive part of the diode voltage drop dissipates power as I^2*R, but for smaller currents, the voltage Vd across the diode is more nearly constant, and the loss is proportional to I*Vd. The total loss is a combination of these two, and is difficult to calculate theoretically. It's probably just as well to resort to measurement.

I set up a variable 12+ volt transformer secondary with a diode bridge and a 15 ohm load. Adjusting the transformer output voltage so that 25 watts were delivered to the load, without any capacitor, the measured diode loss (of one diode) was .605 watts, with an RMS diode current of 570 mA. With a 3000 uF cap across the load, and adjusting the transformer output to again deliver 25 watts to the load, the measured diode loss was .77 watts, with an RMS diode current of 640 mA.

The heating of the diode is more that just "slightly" more in the case of a capacitive load.

With a resistive load, the diode current is not the narrow pulse that it is with a capacitive load, and this results is substantially less diode heating.

 

[colin55]

I was actually considering an old circuit I designed years ago using a power resistor before the capacitor reservoir as the resistive circuit and I did not make myself clear enough. Good you corrected this.

I reckoned the heating would be more with capacitive load but I was not game enough to say it was substantially more.

 

[audioguru]

Thanks.
So i dont understand.
You say that the diode's current can reach even 30A, so its the non-continuous current rating?
But the 1A maximal current rating IS the continuous current rating?

It is spec'd as a power supply rectifier. Its current is allowed to be up to 30A for each very short-duration pulse of current from a transformer to a huge filter capacitor. Then it rests and does nothing in between pulses when it cools. Then the average current is 1A and is allowed if its leads are short enough and are soldered to a pretty big pcb copper area used as a heatsink and if the ambient temperature is at or lower than spec'd on the datasheet (but it will be extremely hot at its max allowed internal temperature).

With the conditions listed above then its continuous allowed current is also 1A.

 

[The Electrician]

It is spec'd as a power supply rectifier. Its current is allowed to be up to 30A for each very short-duration pulse of current from a transformer to a huge filter capacitor. Then it rests and does nothing in between pulses when it cools. Then the average current is 1A and is allowed if its leads are short enough and are soldered to a pretty big pcb copper area used as a heatsink and if the ambient temperature is at or lower than spec'd on the datasheet (but it will be extremely hot at its max allowed internal temperature).

With the conditions listed above then its continuous allowed current is also 1A.

Your description sounds like you're suggesting that when the power supply is in normal operation, and the diodes are carrying a current pulse every 1/60 of a second, the pulses are allowed to be up to 30 amps. If this is what you're saying, it is incorrect.

See the attached scope capture showing diode current and cap voltage at turn on of a power supply. The red trace is the current (at 5 amps per division) through the diodes and the yellow trace is the cap voltage.

When the power supply is first turned on, the filter cap is discharged. There will be a large current pulse (or maybe several) to initially charge the cap to the normal operating voltage. Then the current pulses that occur at 1/60 second intervals during steady state operation will be much less than the initial cap charging pulse(s).

The graph on the 1N4007 spec sheet labeled "Non-Repetitive Surge Current" refers to the initial cap charging surge, not the repetitive pulses that occur when the power supply reaches steady state operation. Before that large (up to 30 amps) surge can be repeated, the power supply has to be turned off and the diodes allowed to completely cool down, at least several seconds I would say.

 

[Willbe]

You also need θja to figure the diode's temp. rise above amb.

 

[audioguru]

I stand corrected (thank you The Electrician).
Philips have a max repetitive current of 10A for their 1N4007 rectifier diode and 30A non-repetitive.

I don't usually operate parts at their absolute max allowed temperature. I don't like being burned or seeing my pcb being charred. So an average of half an Amp is plenty for these little rectifier diodes.

 

[colin55]

I wouldn't trust Philips in anything they say.

I made hundreds of thousands of dollars out of Philips rubbish for more than 15years in the servicing business.
It was always a Philips component that failed. Keep it up Philips!!!!!!

 

[The Electrician]

I don't usually operate parts at their absolute max allowed temperature. I don't like being burned or seeing my pcb being charred. So an average of half an Amp is plenty for these little rectifier diodes.

I concur. I recently bought a low cost (probably Chinese made) bench supply from eBay. It was supposedly rated at 5 amps @ 50 volts. I decided to take a look inside, and the rectifier bridge consisted of 4 1N54xx style diodes, soldered into a phenolic PCB with 1 inch leads. The board for about an inch around where the diodes were soldered in has turned quite dark brown. The unit was represented as new. If the board got that dark just during the manufacturers burn-in, you'd think they would realize they were overloading the rectifiers. Of course, I suppose the supply could have been used, but it didn't look it.

I replaced the leaded diodes with a heat sinkable bridge module.

 

[audioguru]

Diodes rated at (only?) 3A max get pretty darn hot with 5A flowing through them.
The Chinese manufacturer is lucky that nobody knows where to sent their junk for a warranty fix.

 

[colin55]

The Chinese manufacturer simply put together something that an American (or other) designed and had absolutley no idea what he was producing.
The Chinese just copy, but when the product gets tested correctly, it is extremely reliable.
Take TV sets for instance. The first time a TV set fails, it is thrown out.
And the same with kettles, toasters, coffee makers vacuum cleaners etc etc etc.

One school of thought is to leave the leads long so the diode does not de-solder itself. The other way is to solder the diode close to the board and provide a good heatsink. I prefer to do things properly.

 

[ecerfoglio]

Diodes rated at (only?) 3A max get pretty darn hot with 5A flowing through them.
The Chinese manufacturer is lucky that nobody knows where to sent their junk for a warranty fix.

It's a bridge rectifier, so they have 2.5A average current per diode.

Still a little close to the limit, but not a 66% overload

 

[The Electrician]

Have a look at the second page of the GE spec sheet I posted.

It's not enough to know the average current through the diode. You must take account of the conduction angle. If the diode current is a narrow pulse, then the heating of the diode can be much larger than you would think just from the average current.

I may take time to capture the diode current and dissipation under load with the scope and post it here.

 

[Willbe]

Have a look at the second page of the GE spec sheet I posted.

It's not enough to know the average current through the diode. You must take account of the conduction angle. If the diode current is a narrow pulse, then the heating of the diode can be much larger than you would think just from the average current.

I may take time to capture the diode current and dissipation under load with the scope and post it here.

Don't forget you need RMS current to know the heating effect. I think SCR manuals publish RMS vs. conduction angle graphs.

 

[The Electrician]

Don't forget you need RMS current to know the heating effect. I think SCR manuals publish RMS vs. conduction angle graphs.

Knowing the RMS current is not enough to determine the heating effect in a diode. Knowing the RMS current through a resistor completely determines the dissipation, but diodes aren't just resistances; they behave like a (more or less) constant voltage (typically taken to be about .7 volt) in series with some parasitic resistances (silicon, connection to the die, etc.).

The best thing to do is to measure the instantaneous voltage across and current through the diode, multiply the two and average. That's what I'll do with my nifty digital scope.

 

[colin55]

The only thing to do is build the circuit and feel the diode after say 10 minutes.
We had to test a $9,000 design where one of the diodes failed after a few weeks.

The reason was the track-work under the diode was slightly thinner than all the other tracks and this caused the diode to over-heat and fail.
It's that critical.


The expense to replace 300 controllers (they were placed in the frame of a door for a commercial fridge) - $200 (travel expenses and time fitting) for the first door and $50 for each additional door.

The company went under.