I am wondering what the effects are of putting diodes in parallel or putting them in series. (like current capabilities, voltage capabilities etc.) Let's say I have a datasheet of a diode. What characteristics would change ? My estimation is that putting parallel would increase the current capabilities, but may have a negative effect in reverse leakage. I have no idea if I am right or how to test it, so any info on diodes in parallel or series would be great.
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6 Answers
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Connecting diodes in series (AK-AK --|<--|<--) will increase the forward voltage of the resultant diode.
Connecting diodes in series (AK-KA --|<-->|--) will cause an open circuit until peak inverse voltage (smallest diode) is applied on total resultant.
Connecting diodes in parallel (AK/AK --|<-- + --|<--) will increase the current carrying capacity of the diode. See "Current Sharing" document below.
Connecting diodes in parallel (AK/KA --|<-- + -->|--) will not get you a resultant diode conduction in both sides.
Diodes in parallel:
Diodes are frequently connected in parallel in switching power supplies in order to share the current. Here is a document on "Current sharing in parallel diodes".
Thermal Runaway really depends on the diode package and the heat-sink (dissipation) that they are mounted on. The diode in my hand right now has a maximum Tj of 150 C (Vishay STPS30L60CW-N3). Provided enough dissipation in the design, the design can deliver higher current in "diodes in parallel".
answered Jan 7, 2013 at 5:10
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16\$\begingroup\$ I am affraid you cannot increase current capability so easily. At least some small resistors in series would be handy in that parallel combination of diodes. \$\endgroup\$– Al KeppJan 7, 2013 at 6:16
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\$\begingroup\$ "Connecting diodes in parallel (AK/KA --|<-- + -->|--) will not get you a resultant diode conduction in both sides". Incorrect, antiparalling diodes will get you forward conduction, irrespective of polarity, as long as the voltage is sufficient. While one diode is reverse biased, the other one will be forward biased. \$\endgroup\$– BartAug 16, 2017 at 7:14
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\$\begingroup\$ Also, connecting diodes parallel , in order to "increase their current capability" can lead to thermal runaway, destroying the diodes. \$\endgroup\$– BartAug 16, 2017 at 7:17
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\$\begingroup\$ @Bart I have seen diodes connected in parallel in switchers. These are STPS30L60CT diodes connected in parallel in Antec ATX power supply. I believe the design is to increase the output current. \$\endgroup\$ Aug 17, 2017 at 0:19
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Putting diodes in series will add the diode drops together. Reverse leakage (and capacitance) should reduce in this configuration.
In parallel, the drop will stay the same (reverse leakage and capacitance will add), but the current capability may not be much higher, due to the possibility of thermal runaway (since as a diode gets hotter it's Vf drops, then it draws more current relative to the rest, gets hotter still, and so on). You can avoid this somewhat by placing the diodes in thermal contact with each other, and/or using a small resistor in series with each.
answered Jan 7, 2013 at 5:27
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15\$\begingroup\$ The same applies to diodes in series; you cannot rely on increasing the voltage rating unless you add a HIGH value resistance across each diode - rated to conduct say 10x the worst-case reverse leakage current across the volt/temperature range. Otherwise one diode leaks, and the other breaks down... \$\endgroup\$– user16324Jan 7, 2013 at 11:45
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1\$\begingroup\$ @BrianDrummond - Yes, good point, you're right - the figures cannot be accurately predicted. \$\endgroup\$ Jan 7, 2013 at 13:09
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1\$\begingroup\$ What about paralleling 2 diodes that are in the same package? They have the same temperature and same characteristics, so the current is shared roughly equally? I see this often in power supply rectifiers, from people who ostensibly know what they are doing. \$\endgroup\$– endolithMay 12, 2014 at 14:30
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\$\begingroup\$ I would mark this answer as correct. The previous one is very misleading \$\endgroup\$– AK_Aug 28, 2014 at 13:56
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https://www.daenotes.com/electronics/basic-electronics/diode-in-parallel
In a parallel connection, diode with the lowest forward biased voltage drop will try to conduct more current causing Thermal runaway.
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What you say is partially true, but putting two 600v diodes in series does not result in an effective 1200v working, unless you put current balancing resistors across each diode.
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Diodes in series with the same polarity each behave no differently than a single diode. The voltage drop and current capabilities of each diode remains the same. The overall voltage drop of the series combination of the diodes will be equal to the total of all of the diode voltage drops. The current capability of the diodes does not change.
Diodes in parallel with the same polarity each behave no differently than a single diode. However, due to the fact that the current into each diode is lower because of the current divider rule, each diode will have less current flowing through it, and therefore its voltage drop will be lower, as this is a characteristic of diodes. Therefore, assuming the diodes are all very similar in Vf, the overall voltage drop of the parallel combination of diodes will be lower than it would be for a single diode. Although each individual diode's current capability does not change, the parallel combination of diodes can handle more current overall, once again because of the current divider rule.
Great illustration here: http://youtu.be/ZH4fs6xkWbk
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2\$\begingroup\$ If the world was ideal sure, however 2 diodes will never perform exactly the same even if heatsinked together so they won't current share exactly 50/50... meaning the combination of two diodes will always handle less than 100% more current than a single. \$\endgroup\$– cb88Sep 24, 2018 at 15:14
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\$\begingroup\$ Not sure what point you are trying to make. You are correct about "less than 100% more" (obviously) - but I never stated otherwise. \$\endgroup\$– GGBBSep 25, 2018 at 23:48
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\$\begingroup\$ Because someone that didn't know better might assume based on they way you have it written that due to reducing the load on each diode you can actually get to 200%+ load.... instead of a more likely senario of something like 150%-190% depending on how well they are heatsinked together and differences in the diodes themselves. Because "assuming the diodes are all identical" just plain isn't a valid assumption. Also the voltage drop will not be lower... Diodes aren't resistors!? It will be some non linear function of the two, favoring whichever diode has the lowest drop. \$\endgroup\$– cb88Sep 26, 2018 at 14:09
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\$\begingroup\$ I think you need to look into it a bit more. The voltage drop will in fact be lower. Watch the video I linked. As forward current across a diode goes down, forward voltage also goes down (data sheets show this). Instinct might tell you that only one diode can be "on" because it will have a slightly lower Vf, but turn-on is never instantaneous - it is gradual, and I think that's what's at play here. So more than one diode can and will be "on" at a time assuming they are close in Vf - which is what I meant by "identical" (a poor word choice I agree). I've breadboarded this for myself - try it. \$\endgroup\$– GGBBSep 27, 2018 at 23:51
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\$\begingroup\$ That's only if you hold current constant. If you double the current you're back at square one voltage drop wise. And that was my point you'll never get to 100% more current conducted. \$\endgroup\$– cb88Sep 28, 2018 at 14:47
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diodes in series helps for voltage drop but to add resistors across each diode is recommendable in order get the rated drop otherwise the voltage drops would be different because same current will flow through every diode whereas the diodes in parallel helps for getting the required current rating and resistors in series to each diode is recommendable
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1\$\begingroup\$ Why are parallel diodes recommended for series diodes? You may have a point but this needs clarification. One obvious downside to this approach is that more current will flow when the diodes are reverse biased which is less than ideal. \$\endgroup\$ Aug 16, 2017 at 8:05
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2\$\begingroup\$ Welcome to EE.SE. Your answer might be good but since you have used no punctuation or capitalisation it is impossible to know where one sentence ends and another begins. These things matter. -1. \$\endgroup\$ Aug 16, 2017 at 9:07
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