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I have come across the following problem a few times in the past:

You have a break-out board where some pins have already been pulled up or down for convenience, e.g. RESET pulled high to keep the board running even when you don't connect the input pin to your MCU. Now you want the opposite standard behaviour when the pin is not set (e.g. in sleep mode or when no power is applied to the MCU). How can you achieve this? The only answer I can think of is to (physically) remove the pull-up and hook a pull-down between the MCU and the break-out board. I would rather not mess with the SMDs on the break-out board, so is there another solution?

A concrete example is an audio amplifier that is by default on and needs to be pulled down by a pin to switch off. Now when I remove power from the MCU, it automatically turns on again and pops audibly.

Yours gratefully Jerome

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    \$\begingroup\$ Do you mean you want change the default working mode, change the default high to default low without remove the pull-up? \$\endgroup\$
    – diverger
    Commented Nov 6, 2014 at 10:33
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    \$\begingroup\$ A schematic would help us see the problem better. \$\endgroup\$
    – Ricardo
    Commented Nov 6, 2014 at 10:36

3 Answers 3

Reset to default
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The best way is (as you suggest) to remove the pull-up from the PCB and fit a pull-down elsewhere. But, you could try just fitting a pull-down resistor of much lower value than the pull-up. The two resistors will form a voltage divider, but with one resistor much lower in value than the other, the lower value will "win" and pull the pin down. The disadvantage of doing this is higher power consumption then the MCU pulls the pin HIGH. Make sure the pull-down resistor does not draw more current from the MCU pin than it can supply (check the datasheet) when pulling HIGH.

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  • \$\begingroup\$ Thanks, I am going to try that asap. As for current draw, that should not be a problem, the original pull-up is a 100K, so with a board running on 3.3.V the new 5K pullup will draw much less than 1mA. \$\endgroup\$
    – Mampomat
    Commented Nov 6, 2014 at 11:41
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The best is removing the original pull-up or pull-down. If not and if you have enough space, you can add another resistor. There is a rule that the new pulls should be at least 15 time the original pulls. That say, if your pull-up is 15K, then if you want to add another resistor to change it to pull-down, your resistor value should be 1K or so.

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If I read you right, this should work:

WAS is on top, IS on the bottom.

enter image description here

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  • \$\begingroup\$ What would be the advantage of this solution over a simple pull-down with lower resistor values? One of the main aims of switching off the breakout board is to save power. Would your solution be more power-efficient? Cheers \$\endgroup\$
    – Mampomat
    Commented Nov 6, 2014 at 11:45
  • \$\begingroup\$ @MampomatWhat would be the advantage of this solution over a simple pull-down with lower resistor values? Lower power wasted than when pulling up a small resistor. Would your solution be more power-efficient? Depends on how long the small pulldown was pulled up. \$\endgroup\$
    – EM Fields
    Commented Nov 6, 2014 at 12:28

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