Connecting large rectifiers in series and parallel - Page 1
Forum section: electronics / projects

author: woodchips
author_rank: Frequent Contributor
timestamp: November 14, 2016, 10:18:51 am
content: I have a basic AC manual metal arc, or stick, welder.I need to weld some 3/4" steel to a digger bucket. To get maximum heat out of the arc use DC, work positive.I have a pile of IR 220CNQ030 dual Schottky rectifiers recovered from a collection of network racks I bought.Has anyone connected large rectifiers to boost their voltage and current ratings? The welder is 90V 170A so going to need 6 rectifiers in series, 30V each, and two in parallel, 220A per package, 110A per recitifier. There will be four of these stacks of 12 packages to form a bridge rectifier.My main concern is the voltage dividing. Put VDRs or similar across each rectifier, each rated at say 25V. But is there a better solution? Capacitors? Resistors? The welding isn't going to notice a fairly large AC ripple.Forward voltage drops with increase in temperature. So if one package took more than an equal share its disippation will drop forcing other packages to take more current. Seems to be quite a nice automatic negative feedback arrangement.With 48 packages it might of course just be easier to get some higher voltage and higher current rectifiers, just not Sckottky. With non-insulated packages will be quite a problem to mount them all in a box smaller than a holiday suitcase!Comments welcome.

author: oldway
author_rank: Super Contributor
timestamp: November 14, 2016, 11:20:36 am
content: Voltage drop is too high, that's not a good solution.High power bridge rectifiers are cheap.http://www.ebay.de/itm/Vollwelle-Diodenmodul-Einzelphase-Bruckengleichrichter-MDQ-200A-200A-1600V-/122087358538?hash=item1c6cf9404a:g:Q7wAAOSwV0RXsqrg200A 1600V = 23.52 euros.

author: madires
author_rank: Super Contributor
timestamp: November 14, 2016, 12:55:30 pm
content: Quote from: woodchips on November 14, 2016, 10:18:51 amA lower Vf means more current, i.e. it's a thermal runaway. Therefore you would have to add series resistors for proper current sharing. As oldway already mentioned, go for diodes or a rectifier rated for your application.

author: Siwastaja
author_rank: Super Contributor
timestamp: November 14, 2016, 01:53:04 pm
content: Quote from: woodchips on November 14, 2016, 10:18:51 amYou got it exactly the wrong way. It's a self-destructing positive feedback, which is the reason why diodes are difficult to parallel. Lower Vf in the hottest devices quickly causes it to take most of the current, heating it even more. The reduced Vf does not reduce the dissipation as much as the increased current increases it. For example, if two diodes are sharing 10A + 10A at 1.5V and 1.5V, both are dissipating 15W. If one of them drops to Vf=1.4V and there is little resistance in mounting wires, the current sharing can become 5A + 15A, and the devices are now dissipating 7.5W and 21W.Actually paralleling diodes usually involves characterizing and hand-picking parts, good thermal coupling (i.e., mounting the paralleled diodes on a big copper heat spreader next to each other) and some derating of the current per diode. Large modules may use paralleled devices like that internally, and their ratings do include the necessary derating.

author: Someone
author_rank: Super Contributor
timestamp: November 15, 2016, 02:02:25 am
content: As above paralleling diodes is hard. Using them in series is however usually quite easy as the avalanche, reverse leakage, and junction capacitance all help voltage sharing, but you need some headroom and put at least 2-3 more devices than required by the overall voltage standoff.

author: cs.dk
author_rank: Unknown
timestamp: November 15, 2016, 04:45:46 am
content: Which rods will you be using?Many rods are quite capable at AC power.

author: David Hess
author_rank: Super Contributor
timestamp: November 15, 2016, 08:00:02 am
content: I agree with oldway; series connections to raise the voltage rating is going to complicate matters a lot. If you have to do it, I doubt reverse breakdown will be a problem requiring voltage balancing.The diodes are about 200A and 0.5V so a 2.5 milliohm (100 watt!) ballast resistance for each diode may be enough for current sharing.Like oldway again, I think you will be better off buying some cheap but larger diodes even if they are silicon.

author: woodchips
author_rank: Frequent Contributor
timestamp: November 15, 2016, 10:43:50 am
content: Thank you all, and yes, got it exactly wrong on forward voltage! This is why you ask others to design check, something governments seemingly never do!Think I will just have to buy some rectifiers, mounting so many seemed a harder problem than sorting the voltage and current sharing.Where is the piggy bank........

author: oldway
author_rank: Super Contributor
timestamp: November 15, 2016, 02:20:03 pm
content: Power dissipation will be more or less 240W (1.2V x 200A).It will be difficult to find a heat sink big enough to dissipate such a high power.There is an easy solution: buy two of these bridge rectifiers, mount each on his own heatsink and use only 2 diodes of each bridge rectifiers, the two + diodes of one and the 2 - diodes of the other.It is far better than to use the two bridge rectifiers in // because there is no problem of current sharing.

author: Ian.M
author_rank: Super Contributor
timestamp: November 15, 2016, 03:20:41 pm
content: You'd do better to simply string some truck batteries in series rather than trying to convert an AC welder without the right parts.

author: oldway
author_rank: Super Contributor
timestamp: November 15, 2016, 03:35:02 pm
content: You need at least 3 of 4 truck batteries in serie and some kind of current limitation.How much cost one truck battery ?...150...200 US$ ?To buy a 200A MMA/TIG welding DC inverter would be more clever than this.

author: woodchips
author_rank: Frequent Contributor
timestamp: November 15, 2016, 06:51:54 pm
content: Batteries don't work, need some means of current limiting which is easily done with a simple mains transformer/choke arrangement. Also need 70V/90V to strike the arc.I have an inverter based welder, DC, do TIG as well, yet I have never succeeded in actually getting it to weld. Rated at 120A so a 1.6mm rod is well within its capability, but just won't strike. On the other hand I used to have a diesel powered DC welder, 400A, like welding with a stick of rhubarb! That was brilliant, also welded with 1.6mm rods to repair the exhaust pipe on the digger. Sadly long gone.

author: snarkysparky
author_rank: Frequent Contributor
timestamp: November 15, 2016, 07:26:17 pm
content: just preheat that metal with a torch. Lower amps will make doo then.

author: richard.cs
author_rank: Super Contributor
timestamp: November 15, 2016, 10:16:55 pm
content: Batteries certainly do work, and old half-knackered starting batteries are fine. You need the high output current but not much capacity as you can hang a current-limited charger across them and recharge as you go. I used three in series for 36V which was just about enough (24V didn't work at all). Current limiting was a combination of old batteries, cheap jump leads, and about a foot of heavy steel fence wire.

author: langwadt
author_rank: Super Contributor
timestamp: November 15, 2016, 11:15:23 pm
content: Quote from: oldway on November 15, 2016, 02:20:03 pmgetting rid of 240W is a lot easier if you use a more of rectifiers, I don't think decent current sharing istoo hard to achieve if you connect them up with separate wires, 50cm of 1.5mm^2 drops 150mV at 25A