I have a 48v battery powered system is charged by normal lead acid charger that might reach up to 14.5v and maybe 15v per battery ( total 58v to 60v ). The circuit contains components rated for 75v and 100v . As everyone else i get confused by the various ratings and how they are related to my needs:

1. reverse standoff voltage

2. break down voltage

3. maximum clamping voltage

I would like to know which TVS fits 48v circuit (voltage rating wise):

from littel fuse 1.5KE series I cut this image : enter image description here

my first impression is to use the 1.5KE68A because the reverse stand-off voltage is approximately equal to maximum Vcc, but the maximum clamping voltage is way above the 75v rating of the regulator. Even the 1.5ke56 has clamping voltage of 77v.

Question:Which TVS should i choose ? or is this a limitation of TVS and i should consider other type of clamping ?

Note for whom might consider this a duplicate: i check 3 posts on this site but have not found the right answer.

TVS Diode selection

Choosing TVS Diodes

TVS specification understanding

  • 1
    \$\begingroup\$ What is the waveform profile you are being subjected to? \$\endgroup\$ – JonRB Aug 15 '16 at 8:26
  • \$\begingroup\$ @JonRB i donot know the shape and i had a hard time measuring it, but it is caused by a Really Big dc motor. especially at quick stops , or when battery is disconnected . \$\endgroup\$ – ElectronS Aug 15 '16 at 15:08
  • \$\begingroup\$ I don't know for certain, but my gut says a "really big DC motor" would blow these TVSs up with current surge. \$\endgroup\$ – Daniel Aug 15 '16 at 16:18
  • 1
    \$\begingroup\$ @ElectronS. The reason I asked about the waveform is from that you can determine the energy and thus size the TVS. A 2us double exp profile to 750V (5r impedance) would need ~3kW part. I think these tvs will pop for what you want. Consider some L and C instead AND figure out why the motor energy is going in an annoying place \$\endgroup\$ – JonRB Aug 15 '16 at 19:38
  • \$\begingroup\$ @ElectronS Does your motor have a flyback diode? If not, your really big DC motor is acting like a really big inductor, and attempting to switch an active inductor off suddenly leads to a massive and sustained voltage spike as that energy tries to go somewhere. If your motor is unidirectional, a suitably hefty flyback diode may be all you require to mitigate the voltage spike. en.wikipedia.org/wiki/Flyback_diode \$\endgroup\$ – Alex Aug 16 '16 at 4:22

Since OP added information on the actual circuit, here's one solution.

LM5118 is an external switch controller so there's no need to use the same input voltage as you're using for switching. I'd put a 22R series resistor to the switcher input and clamp it with a 68V 5W zener diode. This would comfortably protect the circuit @ 88V input voltage. At least in theory. In practice the zener diode would overheat and die unless you had metal clad PCB or perhaps a solder-on heatsink to dissipate all that heat away. Since the system is battery powered there should be no chance of actually exceeding 60V on steady state operation to start with.

  • \$\begingroup\$ OMG ! how did i not think about that , the input for the controller should not consume current so it can take a 22R in series and a zener or lower voltage TVS diode . GREAT IDEA THANK U \$\endgroup\$ – ElectronS Aug 16 '16 at 9:36
  • 2
    \$\begingroup\$ Sounds like a good idea +1 \$\endgroup\$ – Andy aka Aug 16 '16 at 9:46

This picture should help: -

enter image description here

Stand-off voltage relates to the maximum normal working voltage your system might see and, according to the detail in the question, this will be as high as 60V. However, in your specific situation I would look at the break-down voltage specification.

In the Littel Fuse table it tells you that Vbr is measured at 1 mA and this is a really insignificant current for a large battery so, maybe you could start to consider the 1.5KE56. However, the maximum clamping voltage could be 77 V so this puts in jeopardy components only rated at 75 volts.

The next device down has a clamping voltage of 70.1 volts and this would be suitable for use as a protection device but how much current will it take when 60 V is applied across it?

You cannot make a decision on the information given. Maybe 10 mA leakage current (Ibr) is acceptable and maybe the 1.5KE51 could be chosen through manual select-on-test procedures.

I would consider trying to get components rated at only 75 volts up to a rating of 100 volts.

  • \$\begingroup\$ I have done a test on a TVS 1.5ke56. i can confirm the Vbr rating from 53 to 58v , the current draw was 1-5mA max. at 60 volt it reached 100mA and 300mA at 61.5v the diode got hot . which was tested on a power supply continuously (2-3 seconds) . since i cannot generate transients V clamp cannot be measured . This TVS could work if i want to be on the margin for 2 things : 1- battery charger doesn't output more than 14.5v per cell or 58v total . 2- the absolute maximum rating of regulator is 76v but can take 77v . \$\endgroup\$ – ElectronS Aug 15 '16 at 15:20
  • \$\begingroup\$ this is tough decision , by the way i could not find any 100v replacement for the regulator, it is a buck-boost type wide input type from 3-75v . \$\endgroup\$ – ElectronS Aug 15 '16 at 15:23
  • \$\begingroup\$ Specifically which device and is this the only weak spot? \$\endgroup\$ – Andy aka Aug 15 '16 at 16:14
  • \$\begingroup\$ LM5118 , yes this is the only weak device . \$\endgroup\$ – ElectronS Aug 15 '16 at 17:15
  • \$\begingroup\$ What current is it taking from the supply or what load current does it supply and at what voltage? \$\endgroup\$ – Andy aka Aug 15 '16 at 17:50

If I interpret the OP correctly, you're concerned about your battery charger output stage. That shouldn't require a high-power TVS at all, the regulator chip is responsible for keeping the voltage withing boundary conditions. Main concern would be sparking on connecting the plugs that is not a high energy source. But on TVS chips:

Here's an appnote from Microsemi that explains it in a fairly straightforward manner: http://www.microsemi.com/document-portal/doc_view/14650-how-to-select-a-transient-voltage-suppressor

Simply put, the part you want does not exist. Standoff voltage should be equal or higher than the max normal operating voltage, in your case 60V. So the 64V part is the way to go. On the other hand, max breakdown voltage on that part is 79V which may or may not keep the 75V regulator alive. You would probably get away with using that 58V part but the min breakdown voltage of 64.6V is awfully close to the actual operating voltage.

Finally, the clamping voltage applies to sudden transients, think lightning and other spikes in the power network. Spec says that if you have transient current of 14.8A, the part is guaranteed to keep voltage below 103 volts.

"Other clamping solution" is probably pretty hard to find. You'd need a comparator to trigger the clamping part (probably a mosfet) with enough precision. But how will you power that part? Yes, you could use a 9V zener diode or something but then you're dissipating 51V on the series diode(s)..

  • \$\begingroup\$ TVS is for micro second pulse over voltage induced by lightning. For longer time and constant (not uS transient) protection, you may research if "crowbar" method suits. \$\endgroup\$ – EEd Aug 16 '16 at 4:10
  • \$\begingroup\$ @EEd Yeah, a crowbar usually uses a Thyristor (or maybe a Triac) but these are not very good for DC applications for obvious reasons. Using a comparator + mosfet allows turning the current on and off. Sure there are GTOs but they're not a common part. Also the zeners that could handle the currents here @ 65V are expensive kit. \$\endgroup\$ – Barleyman Aug 16 '16 at 8:35
  • \$\begingroup\$ Crowbar is a concept. Exact implementation varies. It use is more applicable toward high value power supply driving expensive load or when overloading is dangerous to life (says, many kW and above).Concept is (a) minimize chance of mal-function in power supply circuit leading to output over voltage (b) have an independent circuit that 'short circuits' the output to ground and blow fuse in the supply.Fuse is non electronic and high reliability.For reliability,the independent circuit is designed to be simple, small components count so that failure chance, under formal calculation method,is low. \$\endgroup\$ – EEd Aug 16 '16 at 17:16
  • \$\begingroup\$ Crowbar is not used in many power supply. It is needed under above conditions or when 'formal' design reliability target is needed. Says, FAA certification of 1 failure per 'millions' miles. Another concept is Cutoff chip (over and under volt), instead of crowbar, is used in low cost li ion battery protector that is inside cell phone battery. Hope it helps. \$\endgroup\$ – EEd Aug 16 '16 at 17:25
  • \$\begingroup\$ Crowbar is a last defense concept. Normally, the simple circuit will not activate for whole life of the power supply. Once activated, the 'element' needs to have sufficient energy rating to take the current (including filter cap.) until the power supply (high reliability) AC incoming fuse (most up stream as power flow path, so protects everything 'after') blow in xxx ms. \$\endgroup\$ – EEd Aug 16 '16 at 17:32

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