RS485 without a 555

[dhouston]

Hi, I just had a random idea. Ive been using the 555 timer method to automatically switch an RS485 chip from receive to send. I was just thinking though, is the 555 part really needed? Couldnt i just fix the data input pin to high and connect the TX pin from the chip to the pins that change it to a sender (possibly through an invertor). This method would mean it changes to and from sender/receiver faster and requires less components (even if it needs an invertor).

Is there a reason everyone seems to use the 555 method instead of this? Are there any potential problems doing it this way?

The 555 switches from send to receive not receive to send. You could delay 25mS or so before switching the line direction or you could watch the output buffer to see when it is clear (not easy with a software UART) - the 555 does this automatically.

[The Master]

When RE and DE are both low the chip is in receive mode. To send data both pins need to be high then the DI pin also needs to go high. All the 555 circuit does is makes RE and DE high when DI goes high then sets them back to low shortly after DI goes low. So i guess the question is really why is the slight delay needed?

If DI is set to high permenantly then the data pin connects to RE and DE. When data is sent it switches the chip to send mode and because DI is already high it sends a high bit.

Some of my circuits use a 555 and the really small ones use an IO pin to save space. If it could be used how i said then i could save an IO pin or a few components.

[Ioannis]

Do you mean that your data are always 1?

If it is, then sure youdo not need the 555. But what kind of data is always 1? Power supply?

Ioannis

[The Master]

The data isnt always 1. When the TX pin goes low then the RS485 chip resets back to a receiver so the line goes low

[Ioannis]

If the data line retuns to zero,that does not mean end of transmission. Do not get confussed.

End of transmissions happens some time later. This "later" gives the 555 as a monostable multivibrator.

Ioannis

[The Master]

What exactly is this "end of transmission"? I thought it was either on or off

[Ioannis]

I believe there is a problem with the basics here.

E.g. you have to send the following byte:

1 1 1 0 0 1 1 0

You send all the bits and not the '1's only

The RE and DE (if connected together) must be at high level before the first bit and a little after the last bit.

Ifyou send with a baud rate of 9600, then each bit is 1/9600=0,1042msec nd the full 8+2 bits (Start and Stop) will be 1,042msec. So the RE and DE have to be more than 1,042 msec high. Not just for the time that a bit of the data is 1.

Is that more clear now?

Ioannis