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I am new to "electrical engineering" and I am working on a little arduino project. I want to have a SPST switch be used as a digital input on the board.

I have one of the switch leads connected to posative and the other connected to the digital input. The problem with this is that when the switch it off I can't be sure the input will be grounded. How can I make sure the input is grounded when the switch is off?

Would it work if I connected the digital input to a resistor that is connected to ground so when the switch is off it will be grounded, but when the switch is on it wont short out -- would that work?

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3 Answers 3

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The most conventional solution would be to connect one side of the switch to ground. Connect the other to the digital input, and also to a resistor between 1 and 10 K ohms going to the positive supply.

Going the other way, with a pull down resistor as Bruno describes, is possible but less preferred. Many inputs already have a degree of implicit pull-up and will read a '1' if unconnected, though not quite reliably. But if your switch is already connected to the positive rail, then a pull down is an okay solution, though many prefer to use a small resistor when connecting logic inputs to the positive rail.

Many microcontrollers also have internal pullup and/or pull down resistors on GPIO pins which can be enabled by writing to a configuration register. If you are driving an input of such a microcontroller you might not need an external pull up/down resistor at all, though not every microcontroller features these.

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    \$\begingroup\$ Also, if you are using a uC with built in pull ups or pull downs, it's good practice to enable them on pins that aren't being used to keep them in a known state. \$\endgroup\$
    – Matt Young
    Commented Aug 16, 2012 at 18:55
  • \$\begingroup\$ It's actually easier, for my project to connect the switch to ground. So I will try use a pull up resistor. Thanks! \$\endgroup\$
    – Sponge Bob
    Commented Aug 16, 2012 at 19:05
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    \$\begingroup\$ Great! Come back when you want to ask us about "switch bounce" :) \$\endgroup\$
    – gbarry
    Commented Aug 16, 2012 at 21:15
  • \$\begingroup\$ Turns out Arduino will do pullup automatically. I'm not sure how, but in the code there is a Digital "INPUT_PULLUP" and I don't even need a resistor. That was easy. Also, about the switch bounce, I noticed it when I was monitoring the serial output. So I just added a 3ms delay in the code. It saved me another trip to the store. :) \$\endgroup\$
    – Sponge Bob
    Commented Aug 17, 2012 at 16:17
  • \$\begingroup\$ If you wade through the ATMEGA data sheet you will see there is a bit that can be written in a configuration register to enable the pullup. The Arduino libraries are just making it more convenient. For the bounce, ignoring (that) input for a while is a usual course, though 3ms might be a bit short. If the delay blocks your whole system you may need to keep it short, if it only ignores that input than you can make it a bit longer since intentional re-activations would be still slower. \$\endgroup\$
    – Chris Stratton
    Commented Aug 17, 2012 at 16:38
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Yes it would work as you described. That is called a pull-down resistor because it assures that when the contact is open the digital input is at the logic state 0 (low). Usually you can use a 10 KΩ resistor for this purpose.

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That will work, and will give you positive logic: a high level (logic "1") when the switch is closed.

But like Chris says the inverted thing is more common: switch connected to ground, and a pull-up (instead of pull-down) resistor to the power supply. Your logic will be inverted: a logic "1" will correspond with an open switch.

A good reason for the pull-up version is that most microcontrollers have them integrated, and you can enable/disable them depending on your needs. Some microcontrollers also have configurable pull-downs, but these are less common.

If you want an external pull-up 10 kΩ may be a good value. A microcontroller's input can have a leakage current of up to 1 µA, and then 10 kΩ will drop a negligible 10 mV. Lower values are certainly possible, but keep in mind that they will have a larger current to ground when the switch is closed. A 1 kΩ resistor will draw 5 mA at 5 V supply, which is a waste of power really. For the 10 kΩ that's only 500 µA. For very low-power applications you may increase the value to 100 kΩ, but remember the leakage current; 1 µA will give a 100 mV drop!

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