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I have ADM485 already mounted and an output of some device, buffered by CD4050BM. For the circuit to be safe for the RS485 bus, the ADM transiver initially should either be in Z-state or read-state. "Some device" is not always connected the circuit, so I want to use pull-down resistor to archive read-state by default.

Currently I'm not able to post images here, so I'll try to give short explanation of what I'm about to do.

  1. -RE and DE of ADM485 are connected to each other (and this can't be changed in my case)
  2. These both are connected to an output CD4050BM buffer unit (input of one is conected to "some device")
  3. I wish to pull down this junction (-RE, DE, buffer out) with resistor of 10k.

Is it ok to do it this way? Which resistor nominal I should use?

EDIT:

Here is an image

enter image description here

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    \$\begingroup\$ Please add links to the images in comments to your question, someone here will surely edit it in for you. Also, providing links to datasheets of all relevant components is highly appreciated here. \$\endgroup\$
    – Anindo Ghosh
    Jan 8, 2013 at 7:30

1 Answer 1

Reset to default
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If you cannot make sure that your control circuit (e.g. microcontroller) can always keep DE deasserted unless it's actually transmitting (including startup, unprogrammed and similar conditions) you should by all means add a pulldown to DE to avoid tying up the bus. After all, it's about the worst thing that can happen to a RS-485 bus - no other device can communicate in this state. This is how I do it:

A pullup on /RE ensures that no spurious noise is received by the UART (and also allows putting the driver into a sleep mode if that matters). If your /RE and DE are connected together, pull that line in the "receive" direction (down).

Pullup values can vary if you have special requirements but 10K is standard for a digital signal not driven by an open drain.

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  • \$\begingroup\$ Thank you for the detailed answer. My /RE and DE are connected together. So if I pull them down it will not cause conflict with CD4050BM output? \$\endgroup\$
    – no id
    Jan 8, 2013 at 11:00
  • \$\begingroup\$ No it won't, the CD4050 will overpower the pullup without creating a short. However, if the DE/RE line is buffered you probably want the pullup before the buffer (at its input). A schematics of your device would be best. \$\endgroup\$
    – Thorn
    Jan 8, 2013 at 11:21
  • \$\begingroup\$ I have attached an URL of the image that shows the part discussed. Buffer is used to protect device on the left (ARM MCU). So I will unlikely put the pulldown before the buffer. If it is very important, I can try to add one more buffer element, but it will cause some throubles with layout. \$\endgroup\$
    – no id
    Jan 8, 2013 at 11:38
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    \$\begingroup\$ The pulldown after the buffer doesn't make sense because the net it's pulling down is always driven by the buffer. The idea is to keep the signal in defined state when it's not driven by the microcontroller whose pin can be tristated (ex: STM32 pins after reset) so you need to put the pulldown before the buffer. In your current circuit if microcontroller pin is floating the buffer may well be switcing randomly (creating chaos on the bus), and the resistor will have no effect in any case. \$\endgroup\$
    – Thorn
    Jan 8, 2013 at 12:17
  • \$\begingroup\$ Got it. The main idea is to pull down "disconnectable" input. It so obvious now. Thanks a lot for you help! \$\endgroup\$
    – no id
    Jan 8, 2013 at 22:29

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