user832: I have recently designed a shield to my Arduino Duemilanove, which has 8 LEDs, on the digital GPIO, from 0 to 7, but it won't program with the shield on because the voltage drop is too much on the RX and TX lines. My question is: can I have a LED on the TX/RX line with a minimum resistor value that allows me to program the Arduino?

user304: I just found this reference (using ICSP for output as well as programming -- essentially the same thing), and it suggests a 4.7kOhm.
I didn't have a 4.7k handy, and used a 15k instead and it works. 
In my scenario, I have run a sketch with a polarized buzzer connected between RX (digital 0) and GND. When connected directly (with no resistor) the sketch fails to load (but the buzzer works as expected). 
If I put the resistor in series with it, the sketch uploads without fail. 
You also probably realise, but mentioning for clarity/completeness, you will lose all serial port functionality for debugging purposes and won't be able to use Serial.[anything]. Make sure to not even initialise the port. 

832: <p>I have recently designed a shield to my Arduino Duemilanove, which has 8 LEDs, on the digital GPIO, from 0 to 7, but it won't program with the shield on because the voltage drop is too much on the RX and TX lines. My question is: can I have a LED on the TX/RX line with a minimum resistor value that allows me to program the Arduino?</p>
 304: <p>I just found this reference (<a href="http://www.avrfreaks.net/index.php?name=PNphpBB2&amp;file=printview&amp;t=70515&amp;start=0" rel="nofollow">using ICSP for output as well as programming</a> -- essentially the same thing), and it suggests a 4.7kOhm.</p>

<p>I didn't have a 4.7k handy, and used a 15k instead and it works. </p>

<p>In my scenario, I have run a sketch with a polarized buzzer connected between RX (digital 0) and <code>GND</code>. When connected directly (with no resistor) the sketch fails to load (but the buzzer works as expected). </p>

<p>If I put the resistor in series with it, the sketch uploads without fail. </p>

<p>You also probably realise, but mentioning for clarity/completeness, you will lose all serial port functionality for debugging purposes and won't be able to use <code>Serial.[anything]</code>. Make sure to not even initialise the port. </p>
