OK I've seen lots of questions about using parallel rectifiers for increased current, which usually gets the response that they won't conduct equally and are prone to thermal runaway.
I've used (and plan to use again) BYV-32E-200 diodes for a rectifier. These have two diodes in the same package with Cathodes on a common pin. My last power supply I used three for a full wave rectifier, one for the two diodes with the cathodes common, and two (using one only in each package) for the part where the anodes are common (the other diode in each package was shorted).
What I would like to know, in a case where a single diode is more than adequate for the current draw (even the peak) is there any harm (or advantage) in using the two diodes in the package in parallel for each diode in a bridge?
I'm thinking that since they are in the same package they would be thermally coupled as well making them more likely to conduct in a closer to equal fashion.
Note I'm not planning on using any ballast resistors if I do this, if it is necessary to do so I will do it the way I did the last bridge (ie no parallel) as I'd end up needing 8 high wattage resistors per bridge I think
Tony.
I've used (and plan to use again) BYV-32E-200 diodes for a rectifier. These have two diodes in the same package with Cathodes on a common pin. My last power supply I used three for a full wave rectifier, one for the two diodes with the cathodes common, and two (using one only in each package) for the part where the anodes are common (the other diode in each package was shorted).
What I would like to know, in a case where a single diode is more than adequate for the current draw (even the peak) is there any harm (or advantage) in using the two diodes in the package in parallel for each diode in a bridge?
I'm thinking that since they are in the same package they would be thermally coupled as well making them more likely to conduct in a closer to equal fashion.
Note I'm not planning on using any ballast resistors if I do this, if it is necessary to do so I will do it the way I did the last bridge (ie no parallel) as I'd end up needing 8 high wattage resistors per bridge I think
Tony.
As a first guess, paralleling diodes when one will do the job of course changes some things. First, the drop while conducting will certainly be less, as each diode will take some current (not necessarily half). Second, the leakage in reverse will be greater. Third, the capacitance of the junctions will add (not a major concern at power line frequencies but certainly important in switching circuits).
You don't necessarily have to use high power equalizing resistors, depending on how much voltage you drop to swamp the forward drop differences. You can calculate the resistance needed by looking at curves of forward drop and deciding what the variation would be. This is most important at high temperature, where the sharing will be better but the safety margin is less. You could, say, allow nominally 0.1 Volt in the sharing resistors; at ordinary currents this wouldn't dissipate much and will help equalize current a little. It's a trade off.
You don't necessarily have to use high power equalizing resistors, depending on how much voltage you drop to swamp the forward drop differences. You can calculate the resistance needed by looking at curves of forward drop and deciding what the variation would be. This is most important at high temperature, where the sharing will be better but the safety margin is less. You could, say, allow nominally 0.1 Volt in the sharing resistors; at ordinary currents this wouldn't dissipate much and will help equalize current a little. It's a trade off.
See Fig 4. Over a 125C range, at Vf = 1.0V, current varies from 8 to 32A. You will certainly see improvement in conduction loss and heat dissipation capacity (2.4 K/W per diode, 1.6 for both).
You can estimate the differential thermal resistance between the diodes, and estimate a reasonable temperature difference. It's rated as 2.4 K/W resistance individually, but 1.6 in parallel, not 1.2, so there seems to be a series resistance of 0.4 K/W, but that suggests they are 2.0 K/W each, which also isn't right. The actual figure is around 1.7 K/W per diode and 0.7 K/W from common to heat sink (i.e., the copper tab). So for example, if one diode is conducting 20A at 1.0V, and the other is off, the active diode will be hotter by 20*1*1.7 = 34 K, which puts it a lot closer to the other diode on the graph. So their temperatures really won't differ all that much.
In general, put your semiconductors on the same heatsink and don't worry about it. The difference is large for small currents, but these currents are generally less than the individual ratings of the components, so current hogging occurs, but does not matter. At higher currents, the natural resistive component begins to dominate and currents become relatively similar. This is true of diodes, saturated BJTs and IGBTs. I'm not sure how much applies to SCRs, but humongous single SCRs are available (possibly for just this reason, that parallel SCRs are difficult to trigger together), so it's not a problem. (FETs, of course, share by Rds(on). Notice that FETs and BJTs *in the linear range* do not share equally because the current is constant, with no resistive component to even things out. Linear circuits still need emitter resistors or what have you. And vacuum tube circuits always share current nicely, because tubes in spec follow tight tolerances compared to transistors.)
Ugh. t_rr is rated for exactly 1A. I've seen SIXTY AMP diodes with the same rating. The time is absurdly low, no kidding they tested the stupid thing at miniscule current! Why do manufacturers do this?! They should at least some provide a graph of t_rr vs. dI/dt with If as parameter. Those usually have If in useful ranges. And give embarrasingly poor, but realistic, numbers accordingly!
Tim
You can estimate the differential thermal resistance between the diodes, and estimate a reasonable temperature difference. It's rated as 2.4 K/W resistance individually, but 1.6 in parallel, not 1.2, so there seems to be a series resistance of 0.4 K/W, but that suggests they are 2.0 K/W each, which also isn't right. The actual figure is around 1.7 K/W per diode and 0.7 K/W from common to heat sink (i.e., the copper tab). So for example, if one diode is conducting 20A at 1.0V, and the other is off, the active diode will be hotter by 20*1*1.7 = 34 K, which puts it a lot closer to the other diode on the graph. So their temperatures really won't differ all that much.
In general, put your semiconductors on the same heatsink and don't worry about it. The difference is large for small currents, but these currents are generally less than the individual ratings of the components, so current hogging occurs, but does not matter. At higher currents, the natural resistive component begins to dominate and currents become relatively similar. This is true of diodes, saturated BJTs and IGBTs. I'm not sure how much applies to SCRs, but humongous single SCRs are available (possibly for just this reason, that parallel SCRs are difficult to trigger together), so it's not a problem. (FETs, of course, share by Rds(on). Notice that FETs and BJTs *in the linear range* do not share equally because the current is constant, with no resistive component to even things out. Linear circuits still need emitter resistors or what have you. And vacuum tube circuits always share current nicely, because tubes in spec follow tight tolerances compared to transistors.)
Ugh. t_rr is rated for exactly 1A. I've seen SIXTY AMP diodes with the same rating. The time is absurdly low, no kidding they tested the stupid thing at miniscule current! Why do manufacturers do this?! They should at least some provide a graph of t_rr vs. dI/dt with If as parameter. Those usually have If in useful ranges. And give embarrasingly poor, but realistic, numbers accordingly!
Tim
Thanks All sounds like I should give it a go. I modelled in spice and did indeed see slightly less voltage drop with the paralleled diodes. I also realised this morning I could get away with 4 resistors per bridge, I will see whether I can fit them in, obviously the voltage drop on 0.1 ohms will be very small so they probably won't need to be too high a power, again spice should be able to show me what wattage I will need (yes so can ohms law and a few quick calcs but I'm lazy)
I actually will be building two new supplies. The first will only be drawing very small currents (around 200ma) the inrush current for the caps though will be around 26A (well within the peak capabiilty of the diodes). The second however will be for a 100W amp. Still should be well within the capabilities of the diodes but they will be working a lot harder.
Cheers,
Tony.
sounds like I should give it a go. I modelled in spice and did indeed see slightly less voltage drop with the paralleled diodes. I also realised this morning I could get away with 4 resistors per bridge, I will see whether I can fit them in, obviously the voltage drop on 0.1 ohms will be very small so they probably won't need to be too high a power, again spice should be able to show me what wattage I will need (yes so can ohms law and a few quick calcs but I'm lazy) I actually will be building two new supplies. The first will only be drawing very small currents (around 200ma) the inrush current for the caps though will be around 26A (well within the peak capabiilty of the diodes). The second however will be for a 100W amp. Still should be well within the capabilities of the diodes but they will be working a lot harder.
Cheers,
Tony.
Hi Richieboy, I wasn't originally wanting to. Something I saw Jan Didden post was that apart from ensuring even current sharing (which in this case is not really an issue) that by adding the small resistance it will reduce the severity of the charging pulses which may result in smoother ripple (highly paraphrased version from memory).
I just modelled it in spice, and whilst the charging pulse was changed slightly (with 0.2 ohms) it also (as expected) dropped the voltage and overall I don't think that there would be any real benefit in this case. Also the power required even at 0.2 ohms is still quite high, and 2W at least would be required.
I'll just go with parallel without any ballast resistors Just figured it was worth trying if it was going to give any benefit which it doesn't look like it will.
Tony.
I just modelled it in spice, and whilst the charging pulse was changed slightly (with 0.2 ohms) it also (as expected) dropped the voltage and overall I don't think that there would be any real benefit in this case. Also the power required even at 0.2 ohms is still quite high, and 2W at least would be required.
I'll just go with parallel without any ballast resistors
Just figured it was worth trying if it was going to give any benefit which it doesn't look like it will. Tony.
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