TVS in Parallel or Series: Power sharing/handling - Page 1
Forum section: electronics / projects

author: TimNJ
author_rank: Super Contributor
timestamp: December 08, 2022, 02:38:43 am
content: Greetings,Say I needed a really chonky TVS but certain mechanical constraints prevent me from fitting one TVS of appropriate power rating, would you rather use two TVS in series or two in parallel?I opened a commercially available automotive DC/DC converter and found they used two big 5KP type TVS diodes on the inputin parallel, which made me wonder whether a series or parallel arrangement made more sense from a functional/power sharing perspective.In some way I’d think series (1/2 desired voltage rating each) would make more sense. At least current is guaranteed to go through both, whereas in the worst case parallel mismatch, you could have almost all current in one and none in the other.Thanks,TimTim

author: jonpaul
author_rank: Super Contributor
timestamp: December 08, 2022, 10:59:05 am
content: Tim, series only, parallèle will not share current as the curves will never be identical.Lowest Zener or clamp V unit takes all the current.We used Motorola MR2525 or 2535 automotive TAZ for alternator load dump on avaition equipment25 A rated, 20two in series for 28V avaition bus.Jon

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: December 08, 2022, 01:06:15 pm
content: Tim,I think it depends on the nature of transient.A good datasheet, for example Bourns, gives two clamping voltages one for short pulse 8/20us and the other for longer 10/1000us pulses.From the breakdown voltage and clamping voltage you can estimate the slope resistance.For short duration, high-current, applications the slope resistance will dominate and will cause the current to share.For long duration, low-current, application the breakdown voltage will be more important.I am using three SMCJ64A in parallel in a 48V eFuse application. The MOSFET breaks about 90A and the current in the wiring inductance flows into the TVSs. The issue here is not energy it is about peak voltage.Here is a picture:Regards,Jay_Diddy_B

author: TimNJ
author_rank: Super Contributor
timestamp: December 08, 2022, 05:11:43 pm
content: Thank you both.After waking up this morning from a fever dream which involved getting chased down an alleyway by a human-sized DO-201 tranzorb, I had a moment of clarity on this.Basically, for parallel operation, assuming some (inevitable) mismatch between the two, the initial current drawn by the first TVS needs to be high enough that the voltage (across the two) will continue to rise such that the breakdown voltage of the second TVS is reached. Basically, that comes down to source impedance. So, if source impedance is high enough and/or the two diodes are badly mismatched, then the sharing will be poor.I guess you could argue that, for high energy surges conditions (let's say load dump for example), the equivalent source impedancewill bequite low, so the tendency for the parallel diodes to share will be greater.On the other hand, a series arrangement seems pretty much immune to this. Both will (essentially) breakdown at the same time and the percentage sharing is only a function of the breakdown voltage matching, not source impedance, current, and so on.So I tend to agree with Jonpaul that series makes more sense, at least, is more consistent (in theory).A downside of series diodes is higher inductance. I am not sure how important that might be in a general sense. Maybe series diodes will struggle more with transients with higher dv/dt.Some simulations attached.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 08, 2022, 05:20:11 pm
content: Longer duration doesn't even matter that much, because avalanche has positive tempco so they tend to share anyway. It takes some volts for things to get flowing, by which time the other will be somewhere up its slope.But there is one thing you can observe about these kind of situations:- In series, current sharing is perfect, and voltage shares proportionally to device voltage.- In parallel, voltage sharing is perfect; temperature sharing is proportional to the voltage difference.They key is temperature sharing: in parallel, the total equivalent rating will be strictly somewhere between 1 and 2 times the individual rating. It must be more than a single, because there's something else to help out, even if incrementally as it turns out; but it must also be less than double because there will be a difference in threshold voltage and therefore final temperature, and the hottest one must still fall within ratings.I mean, in series there will be some imbalance, by the same reasoning; and PTC works against it, though I would guess by not as much. Still, you can see one will hit Tj(max) a bit sooner than the other, and the total rating will be, at best, somewhat less than N times the individual.Tim

author: TimNJ
author_rank: Super Contributor
timestamp: December 08, 2022, 05:39:14 pm
content: Quote from: T3sl4co1l on December 08, 2022, 05:20:11 pmThanks. Ah, I guess this means long enough duration such that the two can kind of equilibriate (to some degree)? Meaning, for parallel operation, the VZ of the lower VZ diode increases as it warms up, which means the voltage across the two increases such that the higher VZ diode breaks down, and then I could imagine they could sort of come to some equilibrium?Quote from: T3sl4co1l on December 08, 2022, 05:20:11 pmYou mean thebreakdown voltagedifference, right? Obviously, there can be considerable power sharing difference if one never breaks down, so I guess you are talking about difference in VZ, not applied voltage.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 08, 2022, 08:04:37 pm
content: Neat thing is it doesn't actually take time: if it's enough power/energy to raise Tj, it's affecting Vz simultaneously.And if it's not enough energy to raise Tj, clearly it's well within dissipation ratings of just the one part! Nice!And yes, the difference in BV or Vz. Vz should be a sufficient measure as I'd be shocked if the knee has significant shape or slope (or exp rate or whatever) differences between similarly sized parts.(Knee/curve/etc. difference would be very relevant if, say, you wanted to use a MOV in parallel with a TVS for some reason. But, I don't think there'd ever be a reason to do that anyway.)Tim

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: December 09, 2022, 03:04:56 am
content: Hi Tim,I looked at this in LTspice.I found the SPICE models for the 15KPA series TVS diodes on the Littelfuse website.DatasheetCheck the modelsFirst step is to check the models at the datasheet test conditions:The models are accurate.Series vs. ParallelI have put the TVS diodes in a circuit with a load dump generator. The parameters of the load dump generator may need to be adjusted to match your application.The results are clear. The series connection is better. These TVS diodes have a very low slope resistance.I have attached the LTspice models.Regards,Jay_Diddy_B15kpa test load dump.asc(3.63 kB - downloaded 142 times.)tvs data.PNG(182.6 kB. 1330x710 - viewed 2487 times.)

author: Psi
author_rank: Super Contributor
timestamp: December 09, 2022, 04:05:18 am
content: Quote from: jonpaul on December 08, 2022, 10:59:05 amOne exception is when you are trying to protect multiple mosfets in parallel. You may need multiple TVS in parallel physically spread out along the row of mosfets because one TVS in the center might not be close enough to the end mosfets to provide adequate protection

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 09, 2022, 04:48:51 am
content: Yeah, at constant temperature, it'll be pretty poor. Ideally one would find a thermally modeled diode, but I suspect those are rather rare to find. Perhaps a few iterations adjusting TJ until the power dissipation matches the transient thermal impedance?Mmh, good luck finding diodes with thermal impedance curves too, for that matter... Well, Littelfuse at least gives one for that part, but, only down to 10ms? Weird. Well, I suppose for the same pulse width, it's the same point on the curve, so just needs to figure the same C/W, no thermal model needed. So that shouldn't be a big deal at least, just tedious.Tim