Thermal cutoff protection for TVS diode

[andrei_electronics]

Hello,

I need to choose from the market a thermal cutoff to connect it close to a TVS diode, so if the TVS goes too hot, the thermal cutoff interrupts the circuit.

I have to mention that in series with this thermal cutoff it is already a melting fuse and the TVS diode (1.5KE33CA) withstand to peak power pulse up to 1500W. The melting fuse is chosen so its time to trip is shorter than the maximum time allowed for the TVS to withstand peak power up to 1500W.

But in case there is an over voltage around the breaking voltage of TVS, the current will be low and the time to trip of the melting fuse will be longer. This low current, even if it's low (I don't have a specific value) it may heat the TVS diode so after some time it will break the TVS.

Now my concern is how to choose the thermal cutoff, how to choose the temperature for it?

Can you help me with some ways to calculate this temperature for the appropriate thermal cutoff?

Thank you!

 

[jiripolivka]

A diode is heated by current, not voltage. Using a melting fuse is correct to protect your diode from current overheating.
Protecting from an overvoltage is best by design. Simply make sure there cannot be an overvoltage across your diode. Select one with a higher back voltage than might occur in the circuit. If such voltage can exceed say 2-3 kV, use a spark gap for protection. For lower voltages there are gas-discharge overvoltage protectors. Raychem also offers voltage-dependent resistors for such purpose.

 

[andrei_electronics]

A diode is heated by current, not voltage. Using a melting fuse is correct to protect your diode from current overheating.
Protecting from an overvoltage is best by design. Simply make sure there cannot be an overvoltage across your diode. Select one with a higher back voltage than might occur in the circuit. If such voltage can exceed say 2-3 kV, use a spark gap for protection. For lower voltages there are gas-discharge overvoltage protectors. Raychem also offers voltage-dependent resistors for such purpose.

Thank you for the answer.

The thing is that this circuit it is intended to protect from over voltages. So over voltages will appear.

The requirements are established, so that TVS has to be used, the over voltage will occur, and the thermal fuse should be used.

What I need to know is how to calculate the heating in the TVS in order to choose an appropriate thermal cutoff.

Thank you.

 

[jiripolivka]

Thank you for the answer.

The thing is that this circuit it is intended to protect from over voltages. So over voltages will appear.

The requirements are established, so that TVS has to be used, the over voltage will occur, and the thermal fuse should be used.

What I need to know is how to calculate the heating in the TVS in order to choose an appropriate thermal cutoff.

Thank you.

It may be difficult to calculate the parameters. High-current diodes are massive, so it takes some time to heat up over a safe junction temperature. I think you will need to make experiments to find what you need. Junction temperature only depends on forward current.
Backward voltage does not heat the diode, only close to breakdown there is a rise in back current, so then the POWER defined by the high voltage AND backward current product may heat up the junction. Overvoltages are easier to "swallow" by Raychem protectors (blue round blocks in AC/DC converters) or spark gaps. When overheated by the forward current,the breakdown voltage also drops of a P/N junction. I would make experiments, calculations are only good for estimates.

 

[FvM]

What I need to know is how to calculate the heating in the TVS in order to choose an appropriate thermal cutoff.

The TVS diodes have thermal resistance specifications in the datasheet, so you get at least a rough estimation of the case and lead wire temperature for a specific applied power.

 

[andrei_electronics]

It may be difficult to calculate the parameters. High-current diodes are massive, so it takes some time to heat up over a safe junction temperature. I think you will need to make experiments to find what you need. Junction temperature only depends on forward current.
Backward voltage does not heat the diode, only close to breakdown there is a rise in back current, so then the POWER defined by the high voltage AND backward current product may heat up the junction. Overvoltages are easier to "swallow" by Raychem protectors (blue round blocks in AC/DC converters) or spark gaps. When overheated by the forward current,the breakdown voltage also drops of a P/N junction. I would make experiments, calculations are only good for estimates.

Thank you!

- - - Updated - - -

The TVS diodes have thermal resistance specifications in the datasheet, so you get at least a rough estimation of the case and lead wire temperature for a specific applied power.

Thank you!