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| Quick TVS diode question. Vclamp < Vbreakdown |
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| kellogs:
I shall start with, uhm, 10 per year :) Which diode would that be anyway ? I am calculating 4 kJ of energy in that impulse; approximating rise and fall curves with straight lines E = (U^2 / R_i) * 1/2 *(t_r +t_d) Worst case: U = 101V, R_i = 0.5 \$\Omega\$, t_r + t_d ~= t_d = 400ms Now, there is a bit of resistive wiring before 12V hit my PCB as well, and the TVS diode would present some resistance too. I should factor those in too, not sure what values to estimate them at. Does my calculation look right ? |
| T3sl4co1l:
That would be the energy dissipated by the source, assuming it has real resistance, into a short circuit; at most your power is 1/4 that, because of power transfer theorem. But your load or TVS won't be 0.5 ohm: the fact that Vc < Us/2 means E is less as well. If we take more like Vc * (Us/Ri) * td, that's a better approximation of the energy absorbed by the load. So, you can see TVS are superior for having Vc ~ 2 Vnom, whereas MOVs can be more like 3-4x (and they're worse in low voltages, besides). If you're still not convinced that a voltage limiter (switch or regulator type) is viable in many situations... :) Tim |
| kellogs:
Of course! it is V_c * I, not U_s * I... Ok I have found one that I guess would work, although marginally, perhaps not at all if I get unlucky - https://www.mccsemi.com/pdf/Products/15KP17(C)(A)-15KP280(C)(A)(R-6).pdf Extending the P_pp - t_d graph yields some 1kW+ able to withstand over 200ms. Do you know how would a 1A reed switch behave if battered with 2A? Will its life just shorten (to whatever, even 1% will be ok) or will it fail short circuit , or something else? |
| T3sl4co1l:
IIRC, 15kW parts are okay for lower levels of load dump, maybe not the worst case you've selected. There are 30kW parts out there too. The lower voltage rating helps. Note that Fig.1 must eventually flatten out: the 8W continuous rating will take some 10s of s to stabilize at. The sub-sqrt slope of the plot suggests limited thermal diffusion (exactly, diffusion would go as t^(-1/2)). It's not clear if that would get steeper over package-scale time scales (i.e., it takes ~10s ms for heat to diffuse from die to lead frame and package). There's nowhere to go beyond the package, so the curve should steepen at that point, until lead conduction and convection dominate, then finally flatten out at the 8W figure (in the ~10s s range). If we assume the curve remains flat, then we can take its slope of ~ -0.443, and... let's see, the rated pulse is a 10/1000us waveform, but that's the 50% figure; helpfully, the curve shows it reaching 10% at about 4ms. The load dump waveform takes 400ms to reach 10%. If we simply equate these, then we need a 100x time figure, which would be 100^(-0.443) = 7.7 times lower power, or 1.95kW. The load dump peak power is Vc * (Us - Vc) / Ri or 4.8kW, so it seems this part is undersized by ~2.5 times. Which hey, that meshes with my expectation that it's fine for lower ratings, but not the full thing. And in fact an even bigger one is needed then (probably two 30kW in parallel would be safe enough? -- three 15kW would be too marginal I would say, but four or more is probably fine). Contrast with MOVs, for example: https://www.yageo.com/upload/media/product/productsearch/datasheet/cpc/mov/20D_1.pdf Nothing particular about this brand, I think they're all the same materials; ratings/specs are more or less industry standard, curves always look the same. Notice 18-68V types are only rated 20Apk, and the 82V part clamps at the same voltage as the 68V but at 5x the current! (LV MOVs rather stink.) Looking at the clamping curves, the 18V and 22V parts clamp at 100A and 40/50V respectively, which is... useful enough, but the energy ratings are tiny. (A much bigger disc would afford more energy, but likely not enough.) The derating curves suggest for N = 2 about a t^(-1/2) slope, and the pulse is the same type so let's assume 10 times the highest (10ms) point or sqrt(10) less current, or about 13A. Or 22A for the J version ("high surge"). They don't say here BTW, but the maximum current spec I believe is for 8/20us, which, let's see... lines up on the derating plots for single event, yeah. So that's why that figure is so high. So, for about a 100A surge at this level, you'd need more like a stack of 5 of these in parallel, preferably 7-10 for better sharing/reliability. Or larger discs, but you'll still need multiple in parallel I think. Note they aren't rated for leakage; that's probably a bad thing for automotive. As for relays, you're not thinking of using one to disconnect the load, are you? They don't respond fast enough (~1 to 10s of ms). Tim |
| kellogs:
Wow, I need to revisit college notes... >> then we can take its slope of ~ -0.443 How did you derive this number out of that logarithmic graph in fig.1 ? >>The load dump peak power is Vc * (Us - Vc) / Ri or 4.8kW, Of course (2)! Also, reading the note on load dump impulse in my attached picture a few posts back: --- Quote ---a If not otherwise agreed, use the higher voltage level with the higher value for internal resistance, or use the lower voltage level with he lower value for internal resistance. --- End quote --- Does that mean that under worst circumstances, for the diode in question, P_peak yields about 29 * 50 / 0.5 =~ 3 kW ? >>probably two 30kW in parallel Would that not require perfectly matched V_clamp ? These 15 kW diodes are over budget anyway, so, if MOVs are that bad... reed switches! Yes I would actually use a comparator to trigger the reed relay(s). Many of them are specc'ed at 1ms max engage time, which would be great I think. Most of the automotive impulses are ~ uS long, so the reed switch would not even engage them - a smaller TVS diode may be employed for them. And when the big one hits, then TVS takes out a bit of their energy and then 1ms later reed switch relays the rest of it into an 1M resistor or such. But... their carry / switching current ratings are quite lousy. What are they going to do, weld into place ? |
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