I am in the process of designing an input protection circuit for my smart load switch (TPS27SA08)
This circuit is intended to provide reverse polarity protection and overvoltage protection:
The input voltage of the circuit is 12V, 8A is the maximum current.
D1 is a bidirectional TVS diode with 17.1V maximum clamping voltage at 10A and 12V working voltage
Q1 is a P CHannel MOSFET with Vgs maximum voltage +- 20V and Vds of -55V.
The smart load switch voltage input goes from 8-36V
I have 2 questions:
- Do I need to add a fast blow fuse at input for an overcurrent situation or with the tvs is enough?
- Will the MOSFET survive an overcurrent and overvoltage scenario? Keeping in mind the maximum clamp voltage of the TVS and the maximum voltage of Vgs of the MOSFET.
Best Answer
To fully answer this you have to define what the electrical threats are that you are trying to protect your circuit against. For instance, you've used a bidirectional TVS at the input but, what perceived threat is the TVS there to deal with?
You have to decide this. Then you might need to add a fuse or a crowbar circuit or something else. I'd certainly use a protection zener on the gate-source and you don't need anything like a 2 watt resistor feeding the gate.
If it's an ESD threat then certainly you won't need a fuse because ESD events are well defined in energy levels and peak currents and most TVS devices are designed to work with ESD. However, if the "threat" is indirect lightning surges you may well have to fit a fuse should the TVS diode "operate" but fail short circuit. However, the 1N6384 is a beast and can handle a 600 watt "dump" for 10 ms i.e. it certainly looks good for some indirect lightning events. But, the devil is in the detail.
I reiterate - define the threats.