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So I'm building a life-sized battle damaged terminator with a 3.7v 240 mA spy cam in the human eye, and a red 1.8V LED in the exoskeleton exposed eye. When the camera is powered by a 3.7V 340 mA power supply, I want the LED to also be powered but want to make sure I don't overheat or under-power anything. I'm guessing at a parallel(?) circuit like the below:enter image description here

Am I on the right path here or do I also need to up the power supply to account for the 1.8V LED? I understand I only need a 95 ohm resistor but all I have are 100, 220, 330+.

Update: So I found the actual LED Datasheet: enter image description here

Does this mean the LED is actually 2 Vf and 20 mA?

This is my first project and only have worked on basic automotive electronics so my apologies for my ignorance in advance and thank you for taking the time!


UPDATE:

So, turns out the "3.7V" power supplies were actually reading 8V+, so I swapped them out for a standard 5V .5A USB Charging brick, which worked on the camera alone before. Trouble is, with the LED, the Camera would not work, so I tried a 5V 1.0A charger, same problem. So I tried a 5V 2.1A charger and it worked! Trouble is, now the camera overheats. (registering over 40C). So with the revised diagram below, can anyone tell me how I should be able to safely power the camera and LED without overheating the camera while using a basic USB charger? enter image description here

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    \$\begingroup\$ depending on the LED 20mA may be overkill. Double check the datasheet. Most modern LEDs will run plenty bright on 5-10mA. \$\endgroup\$
    – Aaron
    Commented Apr 16, 2020 at 17:20
  • \$\begingroup\$ Thanks Aaron! I did not know that. \$\endgroup\$
    – Boots
    Commented Apr 16, 2020 at 17:52
  • \$\begingroup\$ Was you fV a typo by any chance and supposed to be just V? \$\endgroup\$
    – winny
    Commented Apr 16, 2020 at 19:16
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    \$\begingroup\$ Probably VF. And yes, that means the forward voltage drop of your LED is 2.0 V at 20 mA. $$3.7\ V - 2.0\ V = 1.7\ V$$ $$1.7\ V\div100\ Ω=17\ mA$$ It will probably be very bright depending on the LED...you can go higher on the resistor until it looks right to you. \$\endgroup\$
    – evildemonic
    Commented Apr 16, 2020 at 20:02
  • \$\begingroup\$ The LED is not the reason your power supply could not run the camera. Many of these cameras will overheat even with a suitable voltage power supply, especially the ones which are designed for aircraft and so assume all of a need to minimize weight (ie, no heatsink), airflow, and relatively short duration of operation given flight battery lifetimes. \$\endgroup\$
    – Chris Stratton
    Commented Apr 27, 2020 at 21:25

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All you need to worry about is supplying the correct votlage with a supply that can produce enough current.

A 100 Ω resistor is plenty close enough.

This should work fine.

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