Current through diode.

[rainman1]

I looked at the IF vs. VF of 1N4007 diode:

https://www.tnw.utwente.nl/onderwijs_overig/practica/doc/componenten/1n4007.pdf

I have a few questions please:
1. At the datasheet of 1N4007 it says VF = 1.1V @ IF = 1A. but in the graph, VF = 1.1V @ IF = 2A, is it a mistake? (what is right then?).
2. What happens below 0.6V? (doesnt appear on graph) does it mean that this is the minimal voltage for which the diode will conduct current?

 

[dknguyen]

Below the forward voltage of a diode, the current will decay exponentially. Above this, the voltagea cross the diode will increase very very slowly as the current increases very very quickly (effectively the voltage remains constant regardless of the current).

 

[Willbe]

Diode modelling - Wikipedia, the free encyclopedia

 

[rainman1]

Below the forward voltage of a diode, the current will decay exponentially. Above this, the voltagea cross the diode will increase very very slowly as the current increases very very quickly (effectively the voltage remains constant regardless of the current).

Thanks, what about q. number 1?

 

[dknguyen]

Thanks, what about q. number 1?

THe second half of my answer did answer question one. Above the forward voltage, the voltage changes very little for a large change in current.

 

[colin55]

The 1N4001 to 1N4007 range of diodes are all capable of delivering 1 amp maximum.
In fact they are really only capable of delivering 700mA, as the voltage across them increases as the current increases and this makes them get extremely hot.

I would never allow these type of diodes to pass more than 700mA. Put 2 in parallel or use a 3 amp diode if you need more than 700mA.

If you can't hold your finger on them, they are getting too hot.

 

[rainman1]

Thank you (specially to you Colin, as always!).

So what are you saying?
That everything in the graph that is above 1A is BS?
and thats because that in datasheet it was declared that IF_AV_MAX is 1A?

 

[colin55]

Absolute max of 1 amp. (for short periods of time)

Constant current is 700mA max. Feel how hot the diode gets and work it out for yourself. That's the only way to design things.

 

[rainman1]

Well, i didnt see it was written there that its for short periods of time.
I trust you when you say that on 1A it will get warm, but why arent they mentioning it on datasheet?
Should the graph stop rising at 1A and stay constant on 1A?

 

[colin55]

The data sheets are always hiding the facts of life. That's why you have to ask a person who has already been there, done that, and knows the pit falls.

Power diodes are capable of handling say 1 amp continious and 2 amps for a short period of time and 10 amp for a few milliseconds. That's how they rate them.

This allows a diode to be placed in a power supply and accept current spikes for a short period of time and not get damaged.

The main thing that damages a power diode is a voltage spike. It cannot withstand a voltage spike AT ALL - not matter what the current rating. But it can withstand a current increase of 1,000% for short periods of time.

 

[crutschow]

Thanks, what about q. number 1?

It doesn't say so but the value in the table is probably a maximum value and the values in the graphs are typical.

 

[rainman1]

colin55
Thanks! i learnd a lot.
What about the graph-table collision?
Should IF stay constant above when it reaches 1A, because datasheet says IF_AV_MAX is 1A?

crutschow
Thanks.
As i asked colin (not kelly's old boyfriend from 90210 ) about the graph issue.

 

[The Electrician]

The main thing that damages a power diode is a voltage spike. It cannot withstand a voltage spike AT ALL - not matter what the current rating. But it can withstand a current increase of 1,000% for short periods of time.

This is not strictly true. If a rectifier is to be subjected to reverse voltage transients, then a "controlled avalanche rectifier" can withstand them, within ratings.

See: users.etech.fh-hamburg.de/users/gelab/DownLoad/BY527_1.pdf

or search with google "rectifier, controlled avalanche" for many more examples.

 

[The Electrician]

As i asked colin (not kelly's old boyfriend from 90210 ) about the graph issue.

The graph shows forward voltage for a 300 uS pulse, not a continuous current, so currents well above 1 amp are allowed. See the nearby graph for the "Non-repetitive Surge Current"; currents up to 30 amps are allowed.

Looking at the Absolute Maximum Ratings, you will note that the 1 Amp rating is with .375" lead length. They expect you to solder the leads to some kind of heat sink, such as a relatively large area of copper foil on the PCB, with only .375" of lead between the diode body and the solder joint, and an ambient temperature (the temp of the foil to which the diode is soldered) of no more than 75 degrees C. Under those conditions, the diode can sustain 1 amp average forward current indefinitely.

Also, you said in your first post, "...in the graph, VF = 1.1V @ IF = 2A".

Look at the graph again; it indicates a VF of 1.1V @ IF = 4A.

Crutschow's comment is relevant. The manufacturer didn't indicate which values are typical and which are max or min, so you'll have to draw your own conclusions, but the larger value may well be a maximum.

 

[rainman1]

The graph shows forward voltage for a 300 uS pulse, not a continuous current, so currents well above 1 amp are allowed. See the nearby graph for the "Non-repetitive Surge Current"; currents up to 30 amps are allowed.

Looking at the Absolute Maximum Ratings, you will note that the 1 Amp rating is with .375" lead length. They expect you to solder the leads to some kind of heat sink, such as a relatively large area of copper foil on the PCB, with only .375" of lead between the diode body and the solder joint, and an ambient temperature (the temp of the foil to which the diode is soldered) of no more than 75 degrees C. Under those conditions, the diode can sustain 1 amp average forward current indefinitely.

Also, you said in your first post, "...in the graph, VF = 1.1V @ IF = 2A".

Look at the graph again; it indicates a VF of 1.1V @ IF = 4A.

Crutschow's comment is relevant. The manufacturer didn't indicate which values are typical and which are max or min, so you'll have to draw your own conclusions, but the larger value may well be a maximum.

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?

 

[The Electrician]

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?

When you have a capacitor input filter, which is quite common for small power supplies, the capacitor is uncharged when the supply has been turned off for a while. When you first turn it on, there is a big surge of current which charges up the capacitor.

On the second page of the spec sheet, see the graph titled "Non-repetitive Surge Current"? The graph shows how much current surge the diode can withstand during the initial turn on. "Non-repetitive" means "not continuous"!

The "average rectified forward current" rating in the "Absolute Maximum Ratings" table is for a repetitive half sine wave of current at a 60 Hz frequency. So, the diode rating is for 1 amp average, continuously, with that kind of current waveform.

The Fairchild spec sheet doesn't explain this very well, so I've attached two pages from a GE rectifier spec sheet. On the second page, under the heading "Capacitive loads", the details are explained.

Most capacitor input rectifier circuits won't have a perfect half-sine of current (the current will probably be a more peaked pulse), so the derating factor shown in the chart on that second page will have to be applied.

 

[colin55]

You are still running around in circles.

I said the maximum continuous current for a 1N400x diode is 700mA. I have produced thousands of power supplies and sold thousands of kits using 1N4004 diodes. As soon as the current passes 700mA, the voltage across the diode increases and no matter how large the heasink is, the diodes are far too hot to touch.

 

[The Electrician]

I've attached an image showing the current through the diode in a typical capacitor input filter circuit.

The orange trace is the grid waveform, the purple trace is the waveform at the secondary of the transformer and the blue trace is the diode current.

Just visually estimating I would guess that the conduction angle is somewhat less than 90 degrees. Looking at the derating chart on the second page of the GE spec sheet, we see that the diode average current rating should be reduced to somewhere between 700 and 800 mA.

This is in line with colin55's comment that 700 mA would be an appropriate rating. He was probably considering a diode current that was not a perfect half sine, but was more peaked that that, which is what you'll get with a capacitor input filter circuit.

But the 1 amp rating in the spec sheet is accurate for a purely resistive load, which would have a half sine waveform.

 

[The Electrician]

You are still running around in circles.

I said the maximum continuous current for a 1N400x diode is 700mA. I have produced thousands of power supplies and sold thousands of kits using 1N4004 diodes. As soon as the current passes 700mA, the voltage across the diode increases and no matter how large the heasink is, the diodes are far too hot to touch.

You say the rating is 700 mA; the manufacturer says the rating 1 amp. How do you reconcile the difference? Why would the manufacturer lie to us?

Your empirically determined rating of 700 mA was appropriate because the current waveform was undoubtedly more peaked than a half sine; your power supplies had capacitor input filters, didn't they? The manufacturer's rating of 1 amp is for a half sine, with a conduction angle of 180 degrees.

This is only going to occur for a purely resistive load.

With a capacitor input filter, the waveform will be more peaked than a half sine, and so the diode rating must be reduced as shown in the derating chart on page two of the GE spec sheet I posted.

What was the conduction angle for your power supplies? I suspect that it was similar to the scope capture I posted, and if you referenced your actual conduction angle to the chart in the GE spec sheet, you would find the rating for the diode should be about 700 mA.

There's no mystery here, and no contradiction with the manufacturer's rating of 1 amp in the Absolute Maximum Ratings table. You just have to know under what conditions the manufacturer rates it at 1 amp.

 

[colin55]

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.
The reason is this:
The electrolytic(s) in a normal power supply are charged to say 14v and the energy taken from them during a half cycle will reduce the voltage to say 13.5v
Now the output of the transformer comes along (talking about after the diodes) and it is 10v. Obviously it cannot charge the electrolytic. Even at 13.5v it cannot charge the electro.
It only starts to charge the electro when it is 13.6v. By this time is has nearly reached the peak of its waveform so it doesn’t have much time to charge the electro.
If the average current taken from the power supply is 1amp, the electros must be charged at 4 to 6 amps or more to get the charging done before the waveform starts to fall below 14v.
That’s why the diodes are under great stress.
And that’s why a transformer must be able to deliver a very high current.


These are features that have NEVER been pointed out before.