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Hi Team,
I am using a device from your company in my project namely S25FL512SAGMFIG11
It would be great help if you could share the thermal information of the above device, namely Junction Temperature and the Thermal resistance(Junction-To-Ambient).
Thank you
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Hello Rahul,
Maximum operating junction temperature for the part can be calculated using the below formula.
Tj = Theta JA * P + TA
Tj: Junction temperature
Theta JA: Junction to ambient thermal resistance
P: Power dissipation
TA: Ambient temperature.
For the part S25FL512S ,
P = ICC (max) * VCC (max) = 0.1 * 3.6 = 0.36 W
TA = 85C as it is Industrial grade
Tj = 38 * 0.36 + 85 = 98.68 C.
If you want to calculate at typical values please substitute the current and Voltage values accordingly.
Also refer the application note for more details which explains about the Junction temperature.
http://www.cypress.com/file/210756/download
Thanks,
Krishna.
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Hi,
The thermal resistance value will be 38 C/W.
Thanks,
Pradipta.
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Hi,
The thermal resistance value will be 38 C/W.
Thanks,
Pradipta.
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Thanks,
But what about the junction temperature.
Regards,
Rahul
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Hello Rahul,
Maximum operating junction temperature for the part can be calculated using the below formula.
Tj = Theta JA * P + TA
Tj: Junction temperature
Theta JA: Junction to ambient thermal resistance
P: Power dissipation
TA: Ambient temperature.
For the part S25FL512S ,
P = ICC (max) * VCC (max) = 0.1 * 3.6 = 0.36 W
TA = 85C as it is Industrial grade
Tj = 38 * 0.36 + 85 = 98.68 C.
If you want to calculate at typical values please substitute the current and Voltage values accordingly.
Also refer the application note for more details which explains about the Junction temperature.
http://www.cypress.com/file/210756/download
Thanks,
Krishna.