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So, I've always been using a pull-up resistor in the configuration that usually appears in textbooks but I can't tell why it isn't drawn in another way.

So, basically a pull-up resistor is usually used to prevent a direct connection between ground and the supply voltage when the input pin of some digital device is to be made LOW. The question is why isn't a pull-up resistor drawn as shown by the circuit on the right? In this way, there'll be no loss in voltage from the supply to the pin and there's no fear of short-circuiting the power supply.

two diagrams - on left, pull-up resistor between V+ and pin - on right, pull-up resistor between 0V and the pull-down switch

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    \$\begingroup\$ I think you're switching your outputs between your two examples. In the first (pull up), the driving element is the left node and the output is on the right. In the second (pull down), the driving element is the right node and the output is on the left. But this is all a guess and to get better answers you should label your nodes \$\endgroup\$ – sbell May 11 '15 at 0:35
  • \$\begingroup\$ So, basically a pull-up resistor is usually used to prevent a direct connection between ground and the supply voltage when the input pin of some digital device is to be made LOW. No. A pull-up resistor is to pull the signal HIGH until a direct connection is made to ground (which overrides the high signal set by the weak pull-up). The resistor is used to weaken the high signal so that a direct connection can override it. \$\endgroup\$ – DerStrom8 May 11 '15 at 1:04
  • \$\begingroup\$ Woops, looks like whatsisname beat me to that answer.... :p \$\endgroup\$ – DerStrom8 May 11 '15 at 1:05
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The question is why isn't a pull-up resistor drawn as shown by the circuit on the right?

Because it won't work - the supply is directly connected to the input pin and no matter what you do with the switch, if it has a resistor where you show it to be, it won't change the fact that the input pin is wired directly to the supply.

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  • \$\begingroup\$ I think I missed the fact that for the circuit on the right, the resistor will be in parallel with the impedance of the input pin and so the voltage at the pin is always equal to the voltage across the pull-up resistor which is also equal to the supply voltage. Sorry to ask an obvious question. But what about the size of a pull-down resistor. I mean I've seen a pull-down resistor in the order of 100Ohms used. Doesn't that cause a lot of power dissipation ? \$\endgroup\$ – medwatt Mar 13 '15 at 22:53
  • \$\begingroup\$ @medwatt: 100ohm is definitely overkill for a pullup/pulldown, especially with a CMOS input that has a leakage in the uA. The power loss only happens when the switch is active though. \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 13 '15 at 23:08
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    \$\begingroup\$ @medwatt: It's fairly common to have a 100-ohm or so resistor in series with a switch to protect an input from electrostatic discharge or other such nastiness. A passive pull-up that size would only seem appropriate if the switch was very leaky and e.g. had an "open" resistance of 500 ohms or less and a "closed" resistance of 20 ohms or less. \$\endgroup\$ – supercat Mar 13 '15 at 23:17
  • \$\begingroup\$ It's not great for a pulldown connected to a switch, but for a termination resistor or some situation requiring extremely fast response its reasonable. Or someone left off the "K". \$\endgroup\$ – pjc50 Mar 13 '15 at 23:17
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So, basically a pull-up resistor is usually used to prevent a direct connection between ground and the supply voltage when the input pin of some digital device is to be made LOW

Nope. The purpose of a pull-up/pull-down is nothing to do with circuit protection, and is instead to ensure there is always a valid state on the pin, regardless what is connected.

Without a pulling resistor, when the switch is open, the pin is floating, and might randomly flip between low and high depending on the whims of the electrical fields around you.

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    \$\begingroup\$ For the left image, if there's no pull-up resistor and the switch is open, the output pin will be at Vcc if nothing connects to it. Because there's no current flow through the resistor so it is essentially a wire. \$\endgroup\$ – smwikipedia Jul 1 '16 at 2:52
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For more detailed explanation, please check here. I quoted some below:

Digital logic circuits have three logic states: high, low and floating (or high impedance). The high-impedance state occurs when the pin is not pulled to a high or low logic level, but is left “floating” instead. A good illustration of this is an unconnected input pin of a micro-controller. It is neither in a high or low logic state, and a micro-controller might unpredictably interpret the input value as either a logical high or logical low.

By definition:

Pull-up resistors are resistors which are used to ensure that a wire is pulled to a HIGH logical level in the absence of an input signal.

Pull-down resistors are resistors which are used to ensure that a wire is pulled to a LOW logical level in the absence of an input signal.

So no matter how you draw them, verify your drawings by removing the input signal and see if the logical level is as defined in above quotation.

Below pictures are the illustration of both types.

Note that the switch S1 and S2 serve as the input signal here. They don't have to be SPSTs, other device like BJT transistor can serve that purpose, too, such as here.

Pull-up resistor:

pull-up resistor

Pull-down resistor:

pull-down resistor

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