Diodes ampere rating

[Jackdaniel]

Heya all.. what is "diodes ampere ratings?" for example i got a bunch of 1N4007 diodes and someone said that it is intented to pass only "1.A"! what is that rating? Does that mean it can only pass 1A? does it resist 4A and pass only 1A When i use it to rectify a 5.A transformer?

 

[ericgibbs]

Heya all.. what is "diodes ampere ratings?" for example i got a bunch of 1N4007 diodes and someone said that it is intented to pass only "1.A"! what is that rating? Does that mean it can only pass 1A? does it resist 4A and pass only 1A When i use it to rectify a 5.A transformer?

hi,
Look at this clip from the d/s.
Tell us what you think.

 

[Jackdaniel]

hi,
Look at this clip from the d/s.
Tell us what you think.

sorry eric, but i could't understand it at all, can you pls explain the answer in words? then it would be understandable.. btw the 1n4007 works upto 1000V

 

[ericgibbs]

sorry eric, but i could't understand it at all, can you pls explain the answer in words? then it would be understandable.. btw the 1n4007 works upto 1000V

The Average current [continuous] can be upto 1Amp

A peak current of 30Amps is allowed but only for about 8millisecs.

This applies to all the 1N40xx series

So if you use them for a power supply, its a maximum of 1Amp continuous

 

[Jackdaniel]

ok, thanx eric, and.. One more doubt, what if i connect the 1N4007 to a load that eats 5A current? Can it suppply 5A? Or does it FRY!?

 

[ericgibbs]

ok, thanx eric, and.. One more doubt, what if i connect the 1N4007 to a load that eats 5A current? Can it suppply 5A? Or does it FRY!?

Yes it will die!

 

[colin55]

It will only produce a 700mA DC output. Above this current it gets far too hot.

It's not even a "1amp diode."

 

[Jackdaniel]

It will only produce a 700mA DC output. Above this current it gets far too hot.

It's not even a "1amp diode."

as you said, i was about to power up an amplifier that works at 12V and 5.A, using 1N4007!!. So what item diode should i use to pass 12V 5A? Pls help

 

[colin55]

Use a 5 amp or 10 amp bridge

 

[crutschow]

Using a 5A diode to pass 5A is very marginal and will likely have poor reliability. You should derate at least 50% so I would recommend a 10A diode (or diode bridge) to supply 5A.

And at that current the diodes or bridge will be dissipating at least 3.5W per diode (unless you use Schottky types which would lower the power to perhaps 2W per diode) so they need to be mounted on a heat sink or they will still fry.

 

[carbonzit]

Jack, since your question involves locating actual electronic parts, I'll give you a little help. Let me introduce you to Digi-Key's parametric search facility, which allows you to find parts based on several choices (parameters). (Other vendors have similar search capabilities, but I find that most of them don't seem to work as well as Digi-Key's.)

Here is the starting point if you're looking for rectifier diodes. On this page, try this: Under "Diode Type", select "Standard". Under "Mounting Type", select "Through hole". Last, check the "In stock" checkbox (no point searching for stuff they don't have in stock).

Click "Apply Filters". The next window shows a smaller set of parts. Now you can narrow your search by selecting other parameters, like the current rating. Under "Current - Average Rectified (Io)", try selecting 10A. Click "Apply Filters" again. Now you have a much smaller list which you can look through to find a suitable diode.

 

[MrAl]

Hi guys,


Just a couple notes here about standard rectifier diodes in rectifier applications...


Rectifier circuits are rather unusual circuits even though they seem so simple. The reason is because of the behavior of the diodes under stress with various kinds of post filtering is rather complex. We are dealing with very low impedance devices so there are problems that can come up.

Take a 1 amp diode and pump an already rectified sine wave though it and you get some normal power heating. Double the current but halve the time (half wave rectification) and in basic power heating theory you get to use that diode for twice its rated current. Looking closer however, the diode voltage goes up somewhat and so it causes more power heating than the basic theory would predict. This means we'd have to derate the diode by about 20 percent of twice the rated current (half wave) which brings us to 1.6 amps in this half wave sine wave application, but this is with an actual half wave sine, or a diode in a full wave bridge with resistive load only.

Problems set in however when we add capacitive filtering. Suddenly the diode current is not sine anymore, sometimes no where near sine, but more like a short pulse of current. With only a small amount of capacitor ESR, we could end up with as much as three times the power heating in that diode even in bridge configuration. That's pretty amazing if you ask me, but there is a saving grace built in to most low voltage rectifier circuits: the transformer. The transformer has primary and secondary resistance and even some inductance, which lowers the peak current through the diode and so saves the diode in the end from overheating.

The exact current peak is very hard to calculate without knowing a lot about the transformer, but many transformers have significant equivalent resistance and so brings down the peak current quite a bit, at least in most low voltage low power circuits. In higher power circuits however, we have to be very careful because often the minor resistances are not as large as with the lower power circuits because we dont want excessive inefficiencies to eat up too much of the available input power. We do want to be able to use the diodes to full capacity though, and in rectifier circuits we can sometimes use the diode at a higher than rated current. It depends mostly on the external circuit impedances.

So what's the best medicine here then with all these variables?
It appears that the best is to take a couple measurements and see just what your diodes are doing. Often in a bridge you only have to measure one diode to see what the others are doing too. A measurement of current and voltage on the scope is a good idea, followed by some simple calculations to calculate the power heating. Based on the package size you should be able to determine if you've got a safe workable design or not.

It's interesting that if many of our low voltage circuits that we use every day did not have lossy transformers in them (wall warts) we would probably be burning up diodes on a regular basis

 

[Jackdaniel]

can u give a diode's part number that can pass 10A Current through it?

 

[ericgibbs]

can u give a diode's part number that can pass 10A Current through it?

Look here.
https://www.google.co.uk/url?sa=t&s...pGrWf5lRA&sig2=vdjUfwFPpWtoSOhOBqhglg&cad=rja

 

[RCinFLA]

It's all about how much heating the diode gets and its ability to dissipate the heat. There is a peak current high enough that will blow bonding wires or die connection.

The silicon diode will have a voltage drop depending on current, useable range is usually in 0.6 (low current) to 1.0v(high current). The voltage drop will be lower for a given current as the diode die is bigger and if mounted to a sufficient heat sink.

It can take a large short period current spike since the heating is small over a time average compared to continuous current and its associating heating. The 1N4007 diode voltage drop at 1 amp is 1.1 vdc so it will be dissipating 1.1 watts at continuous 1 amp flow. At this current it will get pretty warm. How warm will depend on lead connection and soldering that aids in heat dissipation.