I need to protect the schematic from the surge transient voltage 1.0 kV 1.2/50 us (standard surge model).

To accomplish this I planned to use two SMBJ200CA TVSs in series at the input and make all following circuit withstanding up to 800V voltage:

enter image description here

I have a filing that this schematic will work fine. However TVS's will work in the unspecified area. Let me explain:

In the datasheet specified that the clamping voltage (Vc) of this part is 324V and the current at this point (Ipp) should be 1.9A:

enter image description here

If I'd look further in the datasheet I will see the voltage-to-current chart ending with Vc/Ipp point:

enter image description here

I checked out Littlefuse, Bourns and SMC Diode Solutions datasheets for the information regarding how it will behave beyond this point - no information :(

I think that 1.9 Amps is definitely not enough for protection. I estimated that the current in the worst case scenario should be up to 500 Amps - 250 times more than specified and I have no source for estimation.

The only thing I can do - put the real device(s) to the test until destruction and made some estimation upon these tests. But I don't like to use any device under unspecified condition as I can not be sure for these specs.

I'd spell the questions in the following way:

  1. Do you share my opinion that this schematic will work fine in real world?
  2. Is TVS a right device for this purpose? Or should I add anything extra to the schematic to make it more reliable (or move working points to the specified area)? I don't like varistors as they are big, expensive and fire-able.
  3. Please share any of your thoughts on the subject...
  • 1
    \$\begingroup\$ Those devices quote a 100A peak surge current on the front page of the datasheet, which might make you more comfortable. It's the power they dissipate which you need to work out. \$\endgroup\$
    – user1844
    Sep 22, 2016 at 12:56
  • 1
    \$\begingroup\$ Add your input side with cable, CM filters and everything to your circuit and the impulse generator's output impedance in order to get any useful simulation result. \$\endgroup\$
    – winny
    Sep 22, 2016 at 13:54

2 Answers 2


1) no, your source is an ideal voltage source. Every source in the real world has some series resistance. So in practice less current will flow but often still enough to blow the TVS.

2) Not on its own. When the TVS reaches its limits, what options does it have ? It can only blow up and then either become an open (so now your circuit will get the overvoltage and be damaged !!!) or become a short, more current will flow, the PCB tracks will vaporize.

You need to do something to make the TVS survive !

Solutions are: input series resistors to limit the current. At the maximum expected voltage the TVS will conduct and the resistors will take all the voltage so make sure they are rated for that.

Another solution is a fuse which will blow and disconnect the input.

The best solution is a combination of a fuse and a resistor. There are also resistors which can act as a fuse.

If you can guarantee that there will only be short overvoltage spikes at the input then the resistor only solution is enough. Fuses are slow so for fast transients they do nothing.

Key point is that you need to dissipate the energy of the spike in a controlled way. Resistors are ideal for this.

  • \$\begingroup\$ I expected that as the spike is short - TVS should be perfect element to dissipate the surge energy (as it developed to withstand the transients). Resistor as energy dissipation point is a bad thing for my opinion. As a current delimiter - I'd agree with you. \$\endgroup\$ Sep 22, 2016 at 11:57
  • \$\begingroup\$ TVS should be perfect element to dissipate the surge energy As long as you do not exceed the max ratings: yes. But if you do exceed the max ratings then there are no guarantees anymore. If series R is not an option then you need beefier devices of better: many in parallel. In that case do watch PCB series resistance so that the load is equally shared. If not then one TVS takes most of the energy and destroys itself. \$\endgroup\$ Sep 22, 2016 at 12:08

You can use a bigger TVS if the duration of your pulse is long. This part is a smallish (600W rating) SMD TVS.

A real current of 500A at (say) 300V clamping voltage = 150kW blow the leads off the part and arc across the molten stubs for the duration of the 'event' so you need to get real on what the spec you are trying to meet actually is.

TVS wattage rating is \$P_{PP}\$ for a certain pulse shape (8/20us or 10/100us) and can be estimated for shorter pulse widths according to the Wunsch-Bell model. You should also derate for for high operating temperature.

A part that can actually absorb a significant duration 150kW pulse will likely be more like the size of a doorknob.

  • \$\begingroup\$ Thank you for such a detailed answer and for the mention of the Wunsh-Bell model: I will get acquainted with it. Regarding the power dissipation: as was mentioned above: resistor put to the very input will significantly limit amount of energy which will be delivered to the whole device (not only to TVS). Even 10 Ohm resistor will reduce the current more than 10 times (in theory). \$\endgroup\$ Sep 23, 2016 at 6:00

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