Understanding RESET-EN Arduino Mega FDTI connections

Hello everyone

I'm trying to create a standalone Arduino board that I want to base on the ATMega2560 and as an interface I want to use the FTDI FT232RL. I've downloaded the schematics of the Arduino Mega since this board incorporated the FT232RL to find out what the connections should be. I've looked at everything and copied it to be able to use it on my standalone board.

The question I have is with regards to the RESET-EN jumper. I don't understand what the use of this is as it is bridged out with another line going to the C12 capacitor going to reset?

From the other older schematics I see that there were resistors used between reset and RTS# and DTR#. Please see the older schematics over here:

So should pin 1 of the jumper on the mega circuit diagram not possibly go to RTS# on pin 3 of the FT232RL? I don't know and hence I'm asking over here. As I look at the jumper in the schematic of the Arduino Mega, it serves no purpose as far as I can understand?

Thank you in advance.


That reset is standard connected to the DTR pin.
Sometimes you do not want the board to reset upon the initialisation of a serial connection.
In that case, you need to cut that connection.
After you've done that, you can fix that cut wire by means of the solder jumper.
That will re-enable the auto reset, and make uploading of sketches a whole lot easier.

In your design, you could just have the jumper and have it preset so uploading will be easy.
After uploading and testing, you could remove the jumper.

If you are not planning on using a serial connection using a USB solution like FT232, this doesn't concern your design.

I don't know why these resistors (of low value) were used.

Hello Mas3

Thank you for your explanation. Unfortunately it still doesn't make sense.

Remember that the jumper in this case is a solder pad jumper and not a removable jumper i.e. it will always be closed unless you go with a carpet knife and specifically cut the track of the jumper between the two pads. So in most cases it will be closed.

Secondly, the way the jumper is bridged out means that reset will always be connected to the DTR pin no matter whether you've cut the jumper or haven't cut the jumper.

Any ideas why it might be wired as such? Thank you in advance.



I know the jumper is shorted by a trace.
That's why i told to cut it (the trace that is, i said connection before) instead of opening the jumper.
You should cut the trace where the schematics show the word 'DTR', that's the only correct spot to do that.
As that is a destructive way of reaching your goal, the design allows to reverse that by offering a solder jumper.
They could have opted for a jumper header (more expensive) or to drop a blob of solder over the jumper, but they didn't (i can see some reasons for all three options).
They even could have opted to shorting the jumper by a trace running through it.

They have implemented a "normally closed" jumper by shorting a normally open jumper footprint with a wire.
You can open the jumper to disable "auto-reset" (with a knife, as you said), and then you can re-close it with a solder-blob or zero-ohm resistor. But the normal state (and the way it usually stays) is exactly as shown in the schematic - the jumper is shorted by the trace.)

Screen Shot 2014-06-10 at 5.37.26 PM.jpg

Hey Guys

Thanks for the update, I understand it perfectly now. Thanks again.