0
\$\begingroup\$

I am looking for advice on protecting my circuit from both transient voltages and reverse polarity. I plan to use a unidirectional TVS diode along with a fuse (fuse directly off the battery, and TVS in parallel with the load), the idea being: normal transients found in an automotive environment will be sent to ground as the TVS clamps, and if the power is connected backwards it will blow the fuse. Then I can easily replace the fuse.

I chose this method due to its simplicity, and because this is a relatively high power circuit (15 A continuous current, fused at 20 A).

The normal operating voltage I expect is around 14.6 V. The lowest rated component in my circuit is an IC with a 40 V rating. As for selecting the TVS voltage ratings - I think this is the easy part. Pick a reverse-standoff voltage around 20 V, breakdown voltage around 25 V, and clamping voltage around 30 V. This gives me some wiggle room on the input voltage, and still protects the IC.

My main concern surrounds the reverse-polarity part. How do I go about selecting the diode power based on the few seconds it will be shorted before the fuse blows? I take it that TVS diodes are not made for sustained current of any kind and might be damaged while it waits for a fuse to blow.

Is there a better, similarly simple way to protect the circuit without introducing any losses or parasitic drains?

\$\endgroup\$
4
  • \$\begingroup\$ What's the part number for your fuse? \$\endgroup\$
    – BeB00
    Jan 1, 2021 at 0:15
  • \$\begingroup\$ There are so many varibles. Type and condition of the battery. length and gauge of wire. Vf vs If at very high If. Etc. Etc. Your best bet is to test it. My guess is that even a 20A fuse will blow in less than 100ms under reverse polarity connection. It is also fairly easy to use the "upside down MOSFET" method of providing reverse polarity protection. Just be mindful to not exceed Vgs maximum on the MOSFET. \$\endgroup\$
    – user57037
    Jan 1, 2021 at 0:23
  • \$\begingroup\$ myelectrons.com/mosfet-reversed-protection See the PMOS version. \$\endgroup\$
    – user57037
    Jan 1, 2021 at 0:39
  • \$\begingroup\$ @BeB00 It's a standard automotive blade fuse - like Littelfuse 0297 series. littelfuse.com/~/media/automotive/datasheets/fuses/… \$\endgroup\$
    – kobra20
    Jan 1, 2021 at 1:10

1 Answer 1

1
\$\begingroup\$

You can just use a normal diode (well not exactly normal but close) in parallel for that, however, it may be better to consider a reverse polarity protection circuit rather than trying to blow a huge fuse.

If your fuse takes a few seconds to blow at say, 50A, thats around 1V * 50A * 3 sec = 150J, which is a pretty large amount of heat. At that level, you will have to consider physical factors like proximity of diode to other components, heatsinking, etc.

This is a simple circuit, there are more complex ones, or ICs that handle it for you. Your high current means you will need a good mosfet, but thats not really a problem (for example).

Another thing you may want to consider is load dumps. These high voltage pulses carry huge amounts of energy, and will explode most TVS devices you would reasonably use. I would do some research into your system to see if these can exist on your device, and look into ISO16750-2. Is this something you're actually going to sell? Or just a one off thing you want to make? If the latter, you might be able to get away with ignoring load dumps.

\$\endgroup\$
8
  • \$\begingroup\$ A 20A fuse may not blow fast at 50A. You are right about that. But a diode forward biased across a car battery will likely result in more amps. Maybe even 200 Amps. That should blow the fuse pretty fast. \$\endgroup\$
    – user57037
    Jan 1, 2021 at 0:20
  • \$\begingroup\$ @mkeith the problem then being that your diode is now dealing with 200A, which is pretty intense. The general issue with this kind of circuit is you dont really have any idea what the upper limit of current is. 1000A from a car battery is certainly achievable, and who knows what would happen in that scenario (the diode would certainly drop more than 1V, for example). How fast would the fuse blow? Would it be fast enough to stop damage to the diode? Probably? Who knows \$\endgroup\$
    – BeB00
    Jan 1, 2021 at 0:25
  • \$\begingroup\$ The higher the current the better in terms of the race condition between fuse and diode. The higher the current the more likely the fuse will blow first. The lower the current, the more likely the diode will blow first. So higher current is good. \$\endgroup\$
    – user57037
    Jan 1, 2021 at 0:29
  • \$\begingroup\$ Personally, I would be OK without the reverse polarity protection (just hook it up right for goodness sake). But MOSFET based reverse polarity is easy to add and should work great. \$\endgroup\$
    – user57037
    Jan 1, 2021 at 0:36
  • \$\begingroup\$ @BeB00 I would like to design everything I make with the intention of selling it whether I end up doing it or not, just so I learn to do it right. Right being - a reasonably efficient method without reinventing the wheel. I have been using a similar circuit with a regular old TVS in parallel without any issues, but have never done a load dump test. I think disconnecting the battery with the engine running is relatively uncommon? \$\endgroup\$
    – kobra20
    Jan 1, 2021 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.