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Question about the L6983I

RFrei.2
Associate

Hello,

Two questions about the L6983I:

-Can I use the L6983I without the isolated output? (basically as the L6983 with higher current output and good availability)

-Why is it limited to 10W? I would want to use it for a 20V output from a 24V rail. With 4.5A output current we would get over the 100W at the ouput here. Is this correct?

Thank you in advance,

Rémi

2 REPLIES 2
Patrizia BELLITTO
ST Employee

Hi @RFrei.2​ ,

thanks for your interest in eDesignSuite and in our products.

About your questions:

  • L6983I without isolated output is just the L6983. Therefore you can use eDesignSuite selecting L6983 instead of the one for L6983I
  • That’s not correct. The 10W are an estimate of the maximum power deliverable at the isolated output. We could generate the 20V (I assume you refer to the primary not isolated output) from 24V input, but we can’t deliver 4.5A for at least two reasons: 4.5A is the peak current limit, not the average; the power dissipation and the consequent thermal shutdown should be taken into account. The max deliverable power is not just the voltage regulated at the output multiplied by the maximum current. If the power dissipation and /or the ambient temperature are too high, the device will be protected by the thermal shutdown protection, as described in the datasheet (https://www.st.com/resource/en/datasheet/l6983i.pdf).

Hope this helps.

Best regards,

Patrizia Bellitto

RFrei.2
Associate

Hello,

Thank you for the answer. This helps a lot. I'll use the design of the L6983 then.

The 4.5A peak is enough in my application.

By the way the L698x serie is one of the only serie of buck converter on the market that i could find that allow to reverse the current in the inductor to actively discharge the ouput capacitor into the input capacitor (when varying the fb resistor value during operation for example). Thank you for that!