Questions regarding diodes in parallel - Page 1
Forum section: electronics / beginners

author: GadgetBoy
author_rank: Regular Contributor
timestamp: June 07, 2018, 07:20:46 am
content: Ok, so I'm working on a Cockroft-Walton ioniser, and I know there some problems with efficiency on NA voltage. So I've been thinking of ways to build a more efficient ioniser, and besides adding more stages (I have a 15 stage one currently, and I'm thinking of adding 5-10 more), I've been thinking about the effects of low-esr caps and changing diode types.So, I was thinking about schottky diodes, and their fast response and low forward voltage, but they're generally rated at no more than 40v.If I put multiple schottky diodes in parallel, will their voltage rating add (i.e. put three 40v schottky diodes in parallel for 120v rating)?

author: Zero999
author_rank: Super Contributor
timestamp: June 07, 2018, 07:59:20 am
content: No, putting diodes in parallel doesn't cause their voltage ratings to add, but remain the same, since the voltage across parallel components is always equal: Kirchoff's voltage law.Connecting diodes in series increases their voltage rating, but it does the same for the voltage drop, so doesn't gain anything efficiency wise.For a relatively high voltage Schottky diode, try the RB168L150.http://www.mouser.com/ds/2/348/rb168l150-774152.pdfThere's also the issue of voltage and current sharing, with the above proposals, because the components aren't perfectly identical.To boost efficiency, cut down on the number of stages and use a transformer.

author: TimFox
author_rank: Super Contributor
timestamp: June 07, 2018, 11:37:09 am
content: When connecting diodes in series, the problem is with the reverse voltage, since the leakage currents are not identical and the voltage will not divide equally across the diodes. Adding high-value shunt resistors across each diode is the usual solution.When connecting diodes in parallel, the problem is with the forward current, since the forward voltages are not identical and the current does not split equally between the diodes. Adding low-value resistors in series with each diode is the usual solution.Both of these solutions are usually a bad idea, compared with buying a single diode with proper ratings.

author: Zero999
author_rank: Super Contributor
timestamp: June 07, 2018, 12:40:07 pm
content: Quote from: TimFox on June 07, 2018, 11:37:09 amYes that's true. Another solution is a two diode package, which won't have the same issues with voltage and current sharing, since the two diodes will be on the same piece of silicon and therefore well matched.

author: David Hess
author_rank: Super Contributor
timestamp: June 07, 2018, 01:23:42 pm
content: Quote from: TimFox on June 07, 2018, 11:37:09 amSome diodes are avalanche rated so this is not a problem for them and they may be used in series without equalization resistors. Usually they are constructed to prevent surface breakdown and to have a uniform junction to prevent partial breakdown which would produce hot spots. High voltage diode stacks take advantage of this but there is no reason separate avalanche rated diodes cannot be used the same way.QuoteThe usual rules about thermal runaway apply so if the forward voltages are matched well enough and the thermal resistance between the diodes is low enough, no series resistors are required. Sometimes I see switching power supply designs which take advantage of this by using both diodes in a dual diode package in parallel.

author: tpowell1830
author_rank: Frequent Contributor
timestamp: June 07, 2018, 04:12:02 pm
content: Quote from: GadgetBoy on June 07, 2018, 07:20:46 amIf you are using a CW multiplier, your diodes do not need to be rated higher than one stage of the multiplier. Best described by Dave, see around 12:30 of video.Hope this helps...

author: ejeffrey
author_rank: Super Contributor
timestamp: June 07, 2018, 04:33:01 pm
content: Quote from: GadgetBoy on June 07, 2018, 07:20:46 amThe way you get good performance/efficiency in a Cockroft-Walton multiplier is to use a high single-stage voltage. In that case, the forward voltage of a shottky diode vs. PN diode doesn't really matter. You are much better off using a diode that can handle 200 V or 500 V reverse voltage and cutting your stages by a factor of 5 or 10 than using 40 V shottky diodes.

author: GadgetBoy
author_rank: Regular Contributor
timestamp: June 07, 2018, 07:24:34 pm
content: Quote from: Hero999 on June 07, 2018, 07:59:20 amI'm feeling very foolish at the moment. I should have remembered our good friend Kirchoff.

author: GadgetBoy
author_rank: Regular Contributor
timestamp: June 08, 2018, 02:20:12 am
content: Quote from: Hero999 on June 07, 2018, 07:59:20 amI've considered winding my own 3:1 transformer for it to kick my 120v mains up to 360. I was envisioning using a capacitive dropper to limit the current into the transformer so I could just whip 30:10 turns on a core of some sort. Would the capacitive dropper keep the mains side of the transformer from melting itself? Or should I just whack a 240v step-up wall wart in the outlet and have done with it?

author: Zero999
author_rank: Super Contributor
timestamp: June 08, 2018, 08:01:46 am
content: Quote from: GadgetBoy on June 08, 2018, 02:20:12 amYou'll need a heck of a load more turns than 30 and if you use a connect a capacitor in series, the capacitor will drop most of the voltage, leaving only a little for the transformer. Any old core is not good enough. It needs to be suitable for mains frequencies and not be too conductive, otherwise there will be huge Eddy losses.Another thing you need to be wary of is, connecting a capacitor in series with an inductor, such as a transformer's primary coil, will actually cause a higher current to flow, than what you'd get from having either the capacitor or inductor alone. The capacitive reactance cancels out the inductive reactance and if they're both the same, you're left with nothing but the effective series resistance of both parts and a very high voltage across both the capacitor and inductor, but opposite in phase. The end result will be damage to the components and hopefully a tripped circuit breaker, otherwise there'll be fire. Look up LRC circuits.Why not simply buy one? Here in Europe, it's possible to get 400V to 110V transformers, which are used to provide 110V for control systems, from a 400V three phase system. All European supplies such as RS Components and Farnell sell them. All you need to do is connect it in reverse: 110V mains to secondary and you'll get 400V.55-0-55V secondary, 400V primary.https://uk.rs-online.com/web/p/toroidal-transformers/3103332/55-0-55V secondary, primary, 208V, to 600V, with plenty of taps, at different voltages.http://uk.farnell.com/block/uste250-2x115/transformer-2-x-115v-250va/dp/1177120

author: GadgetBoy
author_rank: Regular Contributor
timestamp: June 08, 2018, 09:03:51 am
content: Quote from: Hero999 on June 08, 2018, 08:01:46 amUgh, again, now that you've explained to me I realize I should have realized that in the first place. So much for making a compact, low-current mains transformer.