You can get max. RDS on by using the parameter 0.6V Vol @ 8.5mA. The 
result should be 70.5 OHMs when (0.6/8.5) *1000 , the result is 70.5.

70.5 * 21mA = 1470 mV max. So , it's fine if the VF of your LED is 3.5V 
and Vdd = 5V DC if you are willing to drive 3 LEDs by using only one IO.

For series resistor , not necessary or only very small in your 
application. Of course , the max 200mA sink/source per port is the 
limitation.

Thanks &Regards,

Phil Keller wrote:

>Jan-Erik,
>
>  "Very high" could be any value.  Most likely greater than 1 volt at 
>triple the rated current.  If I assume a Vol of 0.2v at 0mA then the Rs 
>is about 47 ohms at 8.5mA.  Thus at 21mA the Vol would be ~1.2v 
>(neglecting other effects.)
>
>  Why do I care?  In this case the LEDs forward drop is rated at 3.5v @ 
>7mA.  That only leave me with about 1.5 volts to play with.  If the 
>output is 0.5v then I need a 140 Ohm resistor but if it is 1v I need a 
>70 Ohm resistor.  If I choose one end or the other I could have a 
>two-to-one variance in LED current and thus about 25% change in 
>luminosity.  Maybe choosing a "middle" value is the best solution.
>
>Phil
>
>Jan-Erik Soderholm wrote:
>
>  
>
>>Phil Keller wrote :
>>
>> 
>>
>>    
>>
>>>The way I am interpreting this is that the device will not 
>>>be damaged at 25mA but the Vol could be very high and
>>>would/could be different from one device to the next.
>>>   
>>>
>>>      
>>>
>>Define "very high" !
>>
>> 
>>
>>    
>>
>>>To guarantee a logic low (Vol <= 0.8) I need to keep the 
>>>current down to 8.5mA.
>>>   
>>>
>>>      
>>>
>>Why does it matter ?
>>Ae you going to use the pin as a LED-drive and as a logic
>>output pin at the same time ?
>>
>>Note also that modern high-efficiency LEDs are
>>perfectly visible down to a few mA's.
>>
>>Regards,
>>Jan-Erik.
>>
>>
>>
>> 
>>
>>    
>>
>
>  
>



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