user16323: I'm trying to make my chinese arduino (which use a chinese CH340G chip) speaks to itself via serial.
I have wired rx/tx together.
Then, I run the arduino IDE and go to the serial monitor.
When I write a message, the arduino should respond with the exact same message, right ? Actually, when I write a message, the arduino doesn't respond anything...

user13432: After some research I found the below schematics of an Arduino UNO with CH340G. A loop-back is not possible as there is a voltage divider (1K ohm, i.e. RX/TX LED) resistors on the RX and TX line.

Tested on a Chinese Arduino Uno clone (VISduino) with CH340G and just as you wrote a loop-back does not work. Tested also an Arduino Mega 2560 clone with CH340G and a Duemilanove clone with FTDI. Both these work just fine. 
Cheers! 

user10794: As Mikael Patel pointed out in his answer, you have set up a voltage divider. Mikael has thoughtfully provided a schematic of the clone Uno, where the relevant part is here:


Equivalent circuit
If you jumper together pins D0 and D1 (Rx and Tx) on the Uno effectively you have made this circuit:


Testing
If I set up this with a function generator, I get this result:

You can see that for a square-wave input of 0 to 5V, we only get an output of 2.16 to 5V.

What is LOW?
Now looking at the datasheet for the DC characteristics we see:

For an input to register as LOW it has to be no higher than 0.3 * Vcc = 0.3 * 5. Thus the voltage cannot be higher than 1.5V. Since we are measuring 2.16V, that will not count as a LOW.
Now admittedly that is for the Atmega328P and not the CH340G, but it will probably have similar characteristics.
I found a datasheet for the CH340G, and that shows even stricter requirements for a LOW signal:


Official Uno
Comparing to the official Uno circuit:

You will note there that the indicator LEDs are on a different circuit, and thus do not form a voltage divider.

The Maths
The clone Uno circuit above can be re-arranged like this:

Given that the LEDs have a voltage drop of 1.86V (measured in practice, and about right for the red LEDs I am using) then effectively we have the resistors R1 and R2 connected to 3.14V.
Since they are in parallel they are effectively 500 ohms.
Now using the formula for voltage dividers:
1000 / (1000 + 500) * (5 - 1.86) = 2.09V

This is close enough to the observed result. I didn't check each resistor to make sure it was exactly in spec.
I don't think R4 affects the results because it is effectively "floating".

16323: <p>I'm trying to make my chinese arduino (which use a chinese CH340G chip) speaks to itself via serial.
I have wired rx/tx together.</p>

<p>Then, I run the arduino IDE and go to the serial monitor.</p>

<p>When I write a message, the arduino should respond with the exact same message, right ? Actually, when I write a message, the arduino doesn't respond anything...</p>
 13432: <p>After some research I found the below schematics of an Arduino UNO with CH340G. A loop-back is not possible as there is a voltage divider (1K ohm, i.e. RX/TX LED) resistors on the RX and TX line.</p>

<p><a href="https://i.stack.imgur.com/Rm50J.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rm50J.jpg" alt="UNO with CH340G"></a></p>

<p>Tested on a Chinese Arduino Uno clone (VISduino) with CH340G and just as you wrote a loop-back does not work. Tested also an Arduino Mega 2560 clone with CH340G and a Duemilanove clone with FTDI. Both these work just fine. </p>

<p>Cheers! </p>
 10794: <p>As Mikael Patel pointed out in his answer, you have set up a voltage divider. Mikael has thoughtfully provided a schematic of the clone Uno, where the relevant part is here:</p>

<p><a href="https://i.stack.imgur.com/YKHdw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YKHdw.png" alt="Clone schematic"></a></p>

<hr>

<h2>Equivalent circuit</h2>

<p>If you jumper together pins D0 and D1 (Rx and Tx) on the Uno effectively you have made this circuit:</p>

<p><a href="https://i.stack.imgur.com/oM9IO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oM9IO.png" alt="Loop-back voltage divider"></a></p>

<hr>

<h2>Testing</h2>

<p>If I set up this with a function generator, I get this result:</p>

<p><a href="https://i.stack.imgur.com/2j648.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2j648.png" alt="Loop-back scope results"></a></p>

<p>You can see that for a square-wave input of 0 to 5V, we only get an output of 2.16 to 5V.</p>

<hr>

<h2>What is LOW?</h2>

<p>Now looking at the datasheet for the DC characteristics we see:</p>

<p><a href="https://i.stack.imgur.com/sNI3Y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sNI3Y.png" alt="DC characteristics"></a></p>

<p>For an input to register as LOW it has to be no higher than <code>0.3 * Vcc = 0.3 * 5</code>. Thus the voltage cannot be higher than 1.5V. Since we are measuring 2.16V, that will not count as a LOW.</p>

<p>Now admittedly that is for the Atmega328P and not the CH340G, but it will probably have similar characteristics.</p>

<p>I found a datasheet for the CH340G, and that shows even stricter requirements for a LOW signal:</p>

<p><a href="https://i.stack.imgur.com/FBn9B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FBn9B.png" alt="CH340G datasheet"></a></p>

<hr>

<h2>Official Uno</h2>

<p>Comparing to the official Uno circuit:</p>

<p><a href="https://i.stack.imgur.com/Oj1bl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Oj1bl.png" alt="Uno schematic"></a></p>

<p>You will note there that the indicator LEDs are on a different circuit, and thus do not form a voltage divider.</p>

<hr>

<h2>The Maths</h2>

<p>The clone Uno circuit above can be re-arranged like this:</p>

<p><a href="https://i.stack.imgur.com/Hv5FO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hv5FO.png" alt="Re-arranged circuit"></a></p>

<p>Given that the LEDs have a voltage drop of 1.86V (measured in practice, and about right for the red LEDs I am using) then effectively we have the resistors R1 and R2 connected to 3.14V.</p>

<p>Since they are in parallel they are effectively 500 ohms.</p>

<p>Now using the formula for voltage dividers:</p>

<pre><code>1000 / (1000 + 500) * (5 - 1.86) = 2.09V
</code></pre>

<p>This is close enough to the observed result. I didn't check each resistor to make sure it was exactly in spec.</p>

<p>I don't think R4 affects the results because it is effectively "floating".</p>
