I'm interfacing with a system using RS-232. To communicate, I use MAX2322 for TTL <-> RS-232 to interface with PIC microcontroller and PL2303 for USB <-> RS-232 to interface with PC.

All of the converters I've used, when not connected to the endpoint, have a symmetric signal on TX pin (+-7, +-6, +-10, depends on the converter). When I connect to the endpoint, the TX pin keeps the voltage for level 1 but the voltage for zero rise 2 or 3V (+7 -4, +6 -3, +10 -7).

That's not a problem (the system still works), but I'm trying to understand why that happend.

The communication uses 4 wires (GND, TX, RX and DTE) and exchange data in a 1200bps rate.

Can you help me understand why that happen?

  • \$\begingroup\$ Just a note: -7V is 1, not 0. \$\endgroup\$ – Johan.A Aug 30 '13 at 6:45

Inside your converter are a USB interface chip and a level converter. The common types of level converters (MAX232 an relatives) use a switched (flying?) capacitor system to create the ~ +/10V from the +5V presented to them. It does this by first converting +5V to +10V, and then converting +10V to -10V. As a result, the -10V supply has a somewhat higher impedance than the +10V supply, so under the same load it will drop more.

Additionally, I am not sure that the load presented by an RS232 receiver is symmetric. The MAX232 datasheet shows a 5k load to ground, but there is no data on the current into the input under various input voltages.

  • \$\begingroup\$ I'm not really into electronics, could you explain me why the -10V higher impedance changes my output signal when I connect to the other device? \$\endgroup\$ – rnunes Sep 23 '11 at 10:25
  • \$\begingroup\$ A practical voltage source can be seen as an ideal voltage source with a series resistor. That series resistor drops a voltage (making the output voltage you see lower), which is proportional to the current you draw. The negative part of your signal is powered from the -10V source, which has a higher 'series resistor'. So, loaded with the same resistor, it will be reduced more than the positive part of your signal (which is powered from the +10V source, which has a lower 'series resistor'). \$\endgroup\$ – Wouter van Ooijen Sep 23 '11 at 10:31

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