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The question is whether DTR pin of FT232RL becomes active only if the chip is plugged into PC USB host where drivers are installed or whenever the chip is powered on by VCC pin? I couldn't find any particular information on this from datasheet.

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  • \$\begingroup\$ Which FT232? There is like 10 different versions, some discontinued. \$\endgroup\$ – Passerby Feb 4 '13 at 22:14
  • \$\begingroup\$ Sorry, I thought DTR logic should be same for all of them. It's FT232RL, I've updated my question. \$\endgroup\$ – Pablo Feb 4 '13 at 22:16
  • \$\begingroup\$ No, because some of the ft232 updates are multifunction chips with multiple modes, not just uart. \$\endgroup\$ – Passerby Feb 4 '13 at 22:35
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Application Note:

AN_184
FTDI Device Input Output pin States
Version 1.0

For Devices: FT232R, FT245R, FT232H, FT2232H, FT4232H & FT2232D

DTR is a Active Low pin. Power on behavior depends on if the chip is set to bus or self powered mode.
Bus powered behavior starts with enumeration in chart, as power comes from the usb host (DTR set to tristate on enumeration, output once active). Since it is active low, I'm assuming it's a voltage high at this point.
Self powered behavior starts at Reset pin being held low (DTR is Tristated). Once the usb connection is added, the USB VDD pulls the reset pin high, and enumeration starts, then active mode.

Suspend refers to Eeprom Setting:

When used in Input Mode, the input pins are pulled to VCCIO via internal 200kΩ resistors. These pins can be programmed to gently pull low during USB suspend (PWREN# = “1”) by setting an option in the internal EEPROM.

enter image description here

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  • \$\begingroup\$ FT232 will be working in USB bus powered mode. So right after FT232 is powered, DTR pin value may change one or more times, correct? The reason I am asking is that I intend to use this pin to control MCU. If device is powered by USB, then MCU will run into bootloader mode, otherwise will boot normally. Basically once MCU gets powered(from USB), DTR supposed to be on its final state. If this can't be answered in short reply, then I might need to open new question. \$\endgroup\$ – Pablo Feb 4 '13 at 23:41

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