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More specifically this RF transmitter which has an incorporated led driver: http://www.mouser.com/ds/2/368/Si4012-35123.pdf

It specifies 4 levels of current on the LED output pin but nothing on voltage overhead. This IC is going to be powered from a 3.3V rail. Will a 1.6-1.8 forward Voltage LED be OK to tie directly to the LED output pin as shown in their diagram ?

5.2.3. PROPERTY: LED_INTENSITY Purpose: LED current drive strength Property: 0x11 Default: 0x00 Fields: LedIntensity [1:0]—LED intensity  00: LED off  01: 0.37 mA  10: 0.60 mA  11: 0.97 mA.

enter image description here

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  • \$\begingroup\$ If it claims to run from down to 1.8V but would not even be able to run a 2V LED when run from 3.6V then I would seriously be worried and not call this an LED driver at all. As you can see it seems to connect the LED to GND as such I would suspect it to be able to just act like any necessary resistor. \$\endgroup\$
    – PlasmaHH
    Commented Feb 28, 2017 at 12:51
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    \$\begingroup\$ Take the minimum supply voltage, say 3.2 V (3.3 V - 0.1 V margin) and 0.5 V drop (my guess) between ground and the LED pin that would leave 3.2 - 0.5 = 2.7 V for a LED. So plenty of margin for a 1.8 V LED I would say. And if it was not OK, then the LED would not light up (as brightly). Then try a different LED. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 28, 2017 at 12:55
  • \$\begingroup\$ @PlasmaHH I am afraid I do not understand. Why would it not be able to run a 2V LED when drawing from a 3.6V source ? Why does it seem to connect LED to GND - from the diagram the LED pin is connected to Vdd through a led diode \$\endgroup\$
    – kellogs
    Commented Feb 28, 2017 at 13:16
  • \$\begingroup\$ @FakeMoustache I am not accustomed to hese LED drivers, should I take it that there is no risk of burning the led ? \$\endgroup\$
    – kellogs
    Commented Feb 28, 2017 at 13:17
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    \$\begingroup\$ should I take it that there is no risk of burning the led Yes you can safely assume that there is no risk for the LED. This chip will simply not provide enough current to damage the LED in any way. Even small LEDs can safely handle 20 mA, this chip only provides a maximum of 0.97 mA to the LED (20 times less). Safe and foolproof, I would say. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 28, 2017 at 13:25

2 Answers 2

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The LED used by SL in their:
i4012 pico board 434 MHz P/N 4012-PSC10B434B (zip contains BOM)
is a Lite-On 1.8V 10mA LED RED CLEAR 0603 SMD:
LiteOn Datasheet P/N LTST-C190CKT

So the LED should be a 1.8V 10ma Red LED

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  • \$\begingroup\$ nono, i understood you well, nice catch! \$\endgroup\$
    – kellogs
    Commented Mar 1, 2017 at 16:04
  • \$\begingroup\$ @kellogs I thought I'd take a quick look for a reference design. Finding an eval board on Digikey was quick. Finding the BOM (usually in manual) was not quick. It became a challenge. SL has very lousy search engine and many boards. \$\endgroup\$
    – Misunderstood
    Commented Mar 2, 2017 at 18:05
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There is no risk of burning a LED. If the rest of the circuitry is fine for you, you should go ahead.

At lower battery voltage, the LED forward voltage may pose a challenge, but, if you limit the lower battery voltage to 3V or higher, everything will be fine. Choose LEDs with lower forward voltage. I am thinking about RED LEDs.

The table described in the datasheet gives you control to limit the current to three different values. The maximum current that can pass through LED is less and there is no harm connecting an LED as mentioned in the application diagram.
If you are concerned, use a current limiting resistor too in series. A value of 100 ohms should come to rescue in case you connect the LED accidentally to ground.

enter image description here

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