Breakdown voltage of ICE3BR2280JZ

[MS_19060803]

Hi all,

Regarding the datasheet Fig.36, the breakdown voltage drops when junction temperature is low.
Could you please tell me the reason?

Best regards,
MS

[Pablo_EG · Accepted solution]

Hello MS-san,

The fact that the breakdown voltage is lower at low temperatures is due to the MOSFET's positive temperature coefficient.
A MOSFET can block more voltage at higher temperatures.
You can think of it as the resistance (@ off state) from drain to source, Rb increasing with temperature.
The higher the Rb for a certain leakage current, the higher the Vb.
The same happens for Rdson, which increases with temperature.

As a deeper explanation, breakdown occurs when carriers gain an ionizing energy in traveling a lattice-scattering mean-free path.
For this, the electric field in the depletion region must reach a critical value.
When increasing temperature, the mean free path decreases, which requires a higher field for the carriers to have enough energy to start impact ionization.
Therefore:
Temperature up => Lower mean free path => Higher electrical field needed = higher voltage needed for breakdown to happen

Best regards,
Pablo

[MS_19060803]

Hi Pablo-san,

 Thank you for your support.

I think that Rb is the resistance shown in the block diagram below.

Is this correct?

Best regards,

MS

[Pablo_EG]

Hello MS-san,

I used the name "Rb" as an example.
Maybe it is better to call it "Rleakage" or "Rdsoff".
The Rb shown in the picture is different.
The conceptual resistance related to the breakdown should be from drain to source.

However, please have in mind that we are using a resistance just to simplify it.
The real phenomenon involves electric fields and carriers, as explained in my previous comment.
Just using the equivalent electrical representation would not be fully accurate.

The answer to the original question is that the CoolSET's breakdown voltage increases with temperature due to the MOSFET having a positive temperature coefficient.

Best regards,
Pablo