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I came acros the following application note for a Depletion Mosfet (https://www.infineon.com/dgdl/Infineon-Application_Note_Applications_for_Depletion_MOSFETs-AN-v01_00-EN.pdf) and i got intrigued. Depletion Mosfet as current source

In the above image, the depletion mosfet is being used as a current source to power a load with a constant current (the best it can, since this type of current source as some oscilations in the output current - from what i read).

My question is, how can you calculate the power that is being dissipated in the depletion mosfet (couldn't find the answer from my search)? It's Vds*Ids?

The objective would be to use the depletion mosfet + a power resistor to dissipate the energy from a set of capacitors (constant current discharge). The objective is that the depletion mosfet dissipates the less possible and that the resistor (load) dissipates the energy.

Best regards

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The short answer is yes, the power dissipated in the MOSFET is Vds * Ids. Assuming that it behaves as a current limiter, the voltage on the resistor would be limited, so limiting the resistor dissipation. At higher voltages than this, the MOSFET would dissipate more power.

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  • \$\begingroup\$ Ok, thanks for your answer. That would be the thing to avoid, the resistor (load) should be the one dissipating that energy, \$\endgroup\$ – dias Dec 14 '18 at 13:16

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