Wikipedia's entry on pull-up resistors states that

A pull-up resistor pulls the voltage of the signal it is connected to towards its voltage source level.

Being that voltage is a measurement and not a physical object, the use of the word "pull" seems to be metaphorical.

However, the sentence also appears to be describing a specific physical phenomenon beyond simply an inexplicable change in a signal's logic level.

What is the physical phenomenon to which the metaphor of "pulling" refers? (If it is a series of events, a description of each event would be most informative.)

  • \$\begingroup\$ Ohm's law, perhaps? \$\endgroup\$ Mar 30, 2017 at 5:32
  • \$\begingroup\$ Thank you for the suggestion. I don't mean to belittle your attempt to help but although I do understand Ohm's law your comment as it is doesn't seem to help my understanding. \$\endgroup\$ Mar 30, 2017 at 5:44
  • \$\begingroup\$ Well, when current flows into the resistor, a voltage difference develops across it. One lead will be at a different potential from the other. If current does not flow into the resistor, both leads are at the same potential. You have transferred the potential at one lead to the other lead. \$\endgroup\$ Mar 30, 2017 at 5:50
  • \$\begingroup\$ ...So, if the 'other' lead is at Vcc, you 'pulled up' the potential; if it is at ground, you 'pulled down' the potential. The switch in the wikipedia figure is there to change the state from no current flows to current flowing. When you close the switch you pass from one state to the other. \$\endgroup\$ Mar 30, 2017 at 5:57
  • \$\begingroup\$ Is it correct that you are saying that to pull the voltage of the signal to which a resistor is connected towards the voltage source level of the resistor is to transfer the potential at the voltage source level to the signal? If so what is the physical process that causes this transfer? (edit: it looks like I wrote this comment before you submitted your second half) \$\endgroup\$ Mar 30, 2017 at 5:57

4 Answers 4


When current flows into a resistor, a voltage difference develops across it, according to Ohm's law. This means that one lead can be at a different potential from the other one.

If current does not flow into the resistor (or if very little current flow, due to the high impedance of the subsequent stage) both leads are at the same potential. You have transferred the potential at one lead to the other lead. So, if the lead with a definite potential is at Vcc, you 'pulled up' the potential; if it is at ground, you 'pulled down' the potential.

The switch in the Wikipedia figure is there to change the state from "no current flows" (hence both leads are at the same potential) to current flowing (hence, the leads can be at different potential). When you close the switch you pass from one state to the other.

By forcing one definite state (usually it's either Vcc or ground), you avoid floating inputs (they are bad because they tend to pick up noise). Also, as Tony pointed out in its answer, you can use an open collector stage as the switch.

But you want to know why, when there is no current flowing through R, both leads are at the same potential. If Ohm's law does not give you a satisfying answer, I guess you have to consider that a conductor where no current is flowing must be equipotential.

Edited to change "other" with "at a definite potential"

  • \$\begingroup\$ From the second paragraph, potential is transferred from "one" lead to the "other" lead. The third paragraph defines "other" as VCC, and from this it seems to follow that potential is transferred from "one" lead to VCC. To a layman like me the word "transfer" suggests proactivity of an actor towards a subject. Is "one" lead acting on the "other"? It seems counter-intuitive that a lead would act on a voltage source. Or is this process a passive one that can be described better with passive verbs instead of active ones like "pull" and "transfer"? \$\endgroup\$ Mar 30, 2017 at 6:33
  • \$\begingroup\$ What I meant is that the potential at the lead in contact with the power supply happens to be present on the unconnected lead as a result of the equilibrium reached in the new configuration. Language can be tricky, especially when used in a colloquial context. You are looking for cause-effect relationships, it seems to me. Well, physics is not much about cause-effect, unless you are able to introduce time explicitly in your model; it's more about relationships between quantities. So, the lead is not causing or acting or doing... it's just there, as are the charges and their associated field. \$\endgroup\$ Mar 30, 2017 at 6:48
  • \$\begingroup\$ Thank you for that last comment; it appears that my confusion comes from the fact that in the context of electronics active verbs are used to describe passive states. This knowledge does help me. \$\endgroup\$ Mar 30, 2017 at 6:54
  • \$\begingroup\$ Separate question, one which will likely be infuriating, but: "When current flows into a resistor, a voltage difference develops across it, according to Ohm's law." Where is that fact visible in V = IR? Again, I may just be getting caught up on language. \$\endgroup\$ Mar 30, 2017 at 7:10
  • \$\begingroup\$ Again, language is - I believe necessarily and unavoidably - imprecise. While I do not believe that V is causing I or I is causing V, based on context I tend to use a verbal form that suggests causation (also, I depicts devices as if they were alive - they are not!). The point is: if you are given definite values of R and I (of system obeying Ohm's law), then the value for V follows from V = R I. If you are given the values of R and V, you can find I. If you are given the values of V and I, then R follows from R = V/I. But this does not mean that resistance is caused by voltage or current. \$\endgroup\$ Mar 30, 2017 at 7:16

Complementary Voltage drivers are called Push Pull drivers.

Since voltage logic normally uses positive voltages,

- PUSH means "away" from ground (0V ="0") and towards (+Vcc ="1" )
- PULL means "towards" ground 

In the case of open collector or open drain outputs the NPN or N-ch FET switches are active switches that PULL towards 0V only and open circuit without any PUSH driver so the positive bias towards Vcc (+) is called PULL-UP ( which is a passive bias R to Vcc or some valid output voltage if it is a high voltage output type. So Active PULL-DOWN or sink and passive PULL-UP or source.

It also applies to high impedance inputs where a logic "1" is needed a pullup to Vcc may be used for many inputs that are not needed to be isolated for test purposes.

Historically logic in TTL used active low for set or reset inputs, so pullup R's were common to disable this feature and up to 10 TTL inputs could be share a Pullup R of 10K. For reasons of avoiding secondary breakdown of reverse biased emitters if V+ happened to exceed max rated Vin on inputs going high these Pullup R's were preferred to limit current. But with CMOS being very high impedance these requirements changed so a direct connection or a pullup R could be used to Vcc if desired or a Pulldown R to 0V with a manual switch to V+ is desired.

  • \$\begingroup\$ I apologize for my lack of basic knowledge but I'm unfamiliar with open collectors, open drain outputs, and NPN or N-ch FET switches. Can this answer be edited so as not to use references to specific components such as these? That may make this explanation clearer for me. \$\endgroup\$ Mar 30, 2017 at 6:07
  • \$\begingroup\$ Oh dear, this answer looks to be far over my head. Perhaps I'd best do more preliminary studying... \$\endgroup\$ Mar 30, 2017 at 6:10
  • \$\begingroup\$ Pullup R just means passive (Resistor) logic = "1" as opposed to active PNP or P-ch driver to logic "1" or PUSH up. Open collector/drain so now you know. Read again until you understand. or read en.wikipedia.org/wiki/Pull-up_resistor \$\endgroup\$ Mar 30, 2017 at 6:14

The metaphor maps electrical quantities and laws to mechanical quantities and laws:

  • voltage ⇆ position
  • current ⇆ force
  • conductance (inverse of resistance) ⇆ spring constant

The relevant laws in both fields are

  • Hooke's law in mechanics: D = F/x
  • Ohm's law (as Sredni Vashtar mentioned) in electricity: R = V/I

According to the metaphor

  • a voltage source is mapped to an upper pixed Point,
  • a resistor connected to this voltage source (e.g. Vcc) is mapped to a spring connected to the fixed point,
  • another resistor connected to another voltage source (e.g. GND) is mapped to another spring fixed to another lower fixed position.

Now depending on spring constant and position of the fixed points both springs pull their common connection point to a certain position.

The same happens electically with the voltage level that is "pulled" up or down by resistors.


In steady state, each signal X can be modeled by a voltage source \$V_X\$ with an internal resistance \$R_X\$. When you add a pull-up resistor \$R_{PU}\$ between X and the power line, the voltage will shift from \$V_X\$ towards \$V_{CC}\$:


simulate this circuit – Schematic created using CircuitLab

$$V = \frac{V_X*R_{PU}+V_{CC}*R_X}{R_{PU}+R_X}$$

Since \$V_{CC}\$ is higher than any signal voltage, adding a pull-up will result in \$V\$ going up, hence the name.

Note that I assumed \$R_X\$ being positive here, which holds true in the vast majority of cases.

  • \$\begingroup\$ Without Rpu is it said that the voltage is pulled by Rx towards Vx? Does this mean that a pin at the point labelled with the voltage V icon is not in a floating state even without Rpu? If so, is this answer also applicable to cases where a pin can float? \$\endgroup\$ Mar 30, 2017 at 8:31
  • \$\begingroup\$ @cheaterpushups A floating pin is a particular case with \$R_X=\infty\$. \$\endgroup\$ Mar 30, 2017 at 8:43

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