Don't worry, stupid questions sometimes turn out to be quite
intelligent. Actually, if you just look at a datasheet you won't
find the answer clearly stated. However, as long as we connect
the bridge as intended, ie. to rectify AC, there will always be
exactly two diodes conducting (or none) and the same current
must flow through both of them. It is thus quite obvious that the
current rating for the bridge is the same as the current rating
for each diode, so there is no need to distinguish these ratings.
Then, the rating given is usually the average forward continous
current. I would think it is always the average, but I am not
sure of this, so better check the datasheet.
intelligent. Actually, if you just look at a datasheet you won't
find the answer clearly stated. However, as long as we connect
the bridge as intended, ie. to rectify AC, there will always be
exactly two diodes conducting (or none) and the same current
must flow through both of them. It is thus quite obvious that the
current rating for the bridge is the same as the current rating
for each diode, so there is no need to distinguish these ratings.
Then, the rating given is usually the average forward continous
current. I would think it is always the average, but I am not
sure of this, so better check the datasheet.
tiroth said:
Normally you have to read the datasheets but sometimes it can be hard to convert data into real life.
To be honest, I'm not totally sure what "Maximum average forward rectified current" means.
I think it is 63% of the peak current. It means 38 A rms on the AC side, assumed resistive load (not an power amp).
Peak current is normally 5-20 times the continious value.
Your exact question: I guess each diode can take at least half the rated current.
Re: Re: Current rating of diodes within a bridge?
Yes you are probably right, I got it wrong. Since it is the average
rectified current and each pair of diodes conducts only every
second half cycle it is more likely that each diode can take half
the rated current.
Peak current is normally 5-20 times the continious value.
Don't you mean peak surge current here?
peranders said:
Yes you are probably right, I got it wrong. Since it is the average
rectified current and each pair of diodes conducts only every
second half cycle it is more likely that each diode can take half
the rated current.
Peak current is normally 5-20 times the continious value.
Don't you mean peak surge current here?
Thanks for the help folks. I was guessing it might be 1/2 of the rated "maximum average" forward current, but after reading a few manufacturer's datasheets I was more confused than when I started.
I wanted to figure out what rating of diodes was necessary to make an equivilent discrete bridge, so I will try using paralleled 8A Schottkeys.
Thanks!
I wanted to figure out what rating of diodes was necessary to make an equivilent discrete bridge, so I will try using paralleled 8A Schottkeys.
Thanks!
Beware that diodes might not share the current equally when
paralleled. I don't know if there is any good way around this.
One might consider small resistors in series with each diode,
but that is hardly a good approach. Maybe you can find
something useful in this application note
http://www.st.com/stonline/books/pdf/docs/3602.pdf
I found it just a few days ago and haven't had time to read it
myself yet.
paralleled. I don't know if there is any good way around this.
One might consider small resistors in series with each diode,
but that is hardly a good approach. Maybe you can find
something useful in this application note
http://www.st.com/stonline/books/pdf/docs/3602.pdf
I found it just a few days ago and haven't had time to read it
myself yet.
That is a pretty useful white paper! However I am not entirely sure how to apply it to the kind of situation commonly seen in charging a large C bank, i.e. very narrow conduction angle.
I guess I will select for lowest possible delta Vf and pray. ^_^ Really these are intentially overspec'd for most designs so I don't anticipate problems.
I guess I will select for lowest possible delta Vf and pray. ^_^ Really these are intentially overspec'd for most designs so I don't anticipate problems.
The diode specs. are given for sine waves, and I suppose the
main limitation is power dissipation, not current, so I would
think they can take quite a lot more current when we have a
narrow conduction angle. We can easily compute the power
dissipation in a diode under the specified conditions. The power
dissipation under the narrow conduction angle might be more
difficult to compute, but to play it safe (I hope) one might
simplify and use the peak current as a constant current drawn
during this whole time period. Then compensate for this
higher current by finding the appropriate voltage drop and
compute this overestimation of the power dissipation. If it is
considerably below the power dissipation under spec'd
conditions and the peak current is not extreme, I would think
this is safe. How many actually do take the narrow conduction
angle into account? My guess would be that most don't or
use som rule of thumb because it seems too difficult.
* Disclaimer: As noted, I don't know for sure if this is safe
and reasonable, it is just my intuition and assesment, for what
it is worth. *
main limitation is power dissipation, not current, so I would
think they can take quite a lot more current when we have a
narrow conduction angle. We can easily compute the power
dissipation in a diode under the specified conditions. The power
dissipation under the narrow conduction angle might be more
difficult to compute, but to play it safe (I hope) one might
simplify and use the peak current as a constant current drawn
during this whole time period. Then compensate for this
higher current by finding the appropriate voltage drop and
compute this overestimation of the power dissipation. If it is
considerably below the power dissipation under spec'd
conditions and the peak current is not extreme, I would think
this is safe. How many actually do take the narrow conduction
angle into account? My guess would be that most don't or
use som rule of thumb because it seems too difficult.
* Disclaimer: As noted, I don't know for sure if this is safe
and reasonable, it is just my intuition and assesment, for what
it is worth. *
Current rating
I think it is not as bad as you think.
35A average current is just that, 35 amps average. (Wait, don't shoot me yet). So say if you build a ps to power a class A amp with 3A standing current, that is the average curent taken from the supply (x 2 for stereo). Your 35 amps bridge will have no trouble whatsoever with this. The limiting factor will most probably be the max repetitive peak current, normally specified for 50 or 60 Hz. That depends on the filter electrolytics as already noted above. Stay reasonabel here and you'll be fine.
Paralleling diodes is quite hard to do without wasting precious power and is totally unnecessary.
Jan Didden
I think it is not as bad as you think.
35A average current is just that, 35 amps average. (Wait, don't shoot me yet). So say if you build a ps to power a class A amp with 3A standing current, that is the average curent taken from the supply (x 2 for stereo). Your 35 amps bridge will have no trouble whatsoever with this. The limiting factor will most probably be the max repetitive peak current, normally specified for 50 or 60 Hz. That depends on the filter electrolytics as already noted above. Stay reasonabel here and you'll be fine.
Paralleling diodes is quite hard to do without wasting precious power and is totally unnecessary.
Jan Didden
Well, I already have a goodly quantity of TO-220 8A dual Schottkey-type rectifiers, so that is where the paralleling came in. I think I paid something like US$0.05 per piece surplus which is quite a lot less than new!
Unfortunately I have been unable to locate a datasheet so I was going to parallel as a safety measure.
Unfortunately I have been unable to locate a datasheet so I was going to parallel as a safety measure.
tiroth said:
I tried Google.
http://pubpages.unh.edu/~aperkins/pdf/Misc/crossref.pdf
Maybe it can be at some help. GI = General Instruments if I'm not wrong?
janneman said:
100 watts @ 4 ohm x 2 (stereo) = 10 A rms = 5 A from each rial => approx 10-15 A peak (due to smoothing caps) all the time. I think the "headroom" for the diodes is a little small.
It's maybe OK for normal music but for party use?
I think 15-25 A bridge is good choice for medium amps. Why save money on such important part which is rather cheap?
peranders said:
5A from each rail for continuous 100W @ 4 Ohms? Must be some party! (Must be very special music also).
Even then, there is a 60% safety margin. And we all know that the real average is closer to 10 or 20W, which is just a couple of amps. So you get closer to 400% safety margin.
And the peak value is probably higher than you say, but that is no problem. repetitive peaks of 5 or 10 times the average spec is not an uncommon spec for these diodes.
And yes, a 15-15 amp is a good choice, but in this case the guy has the diodes laying around. So, you ask the wrong question: "Why save money on such important part which is rather cheap?" The real question is: "Should I spend more money without benefit because I already have what I need?" Answer: No.
Jan Didden
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