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Looking at the datasheet of H11L1 Datasheet, for the Total Device, the power dissipation is 250 mW, does this mean that assuming I put 5V (Arduino) on it, it uses 250 mW / 5 V = 50 mA ?

I think it's quite a lot (assuming I need 3 or 4 in my enclosure), that alone is already 150-200 mA.

(Also I'm wondering if very basic components like diodes, capacitors between GND/5V of ICs etc), pullup/down resistors also uses a lot of power?

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    \$\begingroup\$ That is an absolute maximum value, check page 2 for actual supply current values or figure 6. on page 3. \$\endgroup\$
    – Bence Kaulics
    Commented Jun 2, 2017 at 10:43
  • \$\begingroup\$ @BenceKaulics There I see a supply current of ' Typ' 1.6 mA and 'Max' 5 mA, that looks much better, thanks! \$\endgroup\$
    – Michel Keijzers
    Commented Jun 2, 2017 at 10:48

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This 250 mW is an absolute maximum rating which means that if something goes wrong you can be sure that the device won't blow up if the power dissipation is below this value. This is far above the normal operational conditions.

What you need is figure 6. from page 3.

enter image description here

A couple of mA that is normally needed.

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    \$\begingroup\$ This is only the supply current into the VCC pin. There is also a current into the output pin when the output is active, and that must be included into the allowed power dissipation; the datasheet cannot say anything about it because it's determined by the external pull-up resistor. \$\endgroup\$
    – CL.
    Commented Jun 2, 2017 at 11:20

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