It is not called a pullup in that case. It would just be a current limiting resistor for relay coils, for example. Pullup means specifically for logic where if no switch/transistor is conducting to pull an output low, thenzero current flows through the resistor so no voltage drop therefore the voltage on both ends of the resistor is the same (and one end is the output while the other end is connected to +V, so output becomes +V.

Connecting the gate of a transistor to a voltage higher than it is rated for, even through a resistor, will likely damage it, because the failure mode is due to junction breakdown voltage not an over current phenomenon.

I can understand the logic low/high/floating issue with inductance and so on. But, in this example, it doesn't seem to apply. And, if I remove the "pull up resistor", it seems like output would still be V+, when T2 and/or T3 are "off". So if the resistor is not there to prevent too much current, what is it doing? electronics.stackexchange.com/questions/586734/…

I've looked through a lot of content, and, the "it avoids shorts" keeps recurring. Like this guy here says, youtu.be/u3Xiy2DVnI4?t=59. Seems to be more about not wasting electricity than protecting parts, maybe that's where my assumption was wrong? [Edited by a moderator.]

According to that guy, the technical definition of "pull up resistor" is that the lower the resistance, the stronger it is. And the reverse, a very strong resistor is a weak pull up, acting almost like an open switch and causing floating logic issue. And, that no resistor, is the strongest "pull up resistor". But that it causes short circuiting, and that this is why some resistance is needed.

A pull-up is usually used when a (digital) input needs a defined value when it’s not driven by any other signals. For example when a button is connected to “ground” on one side but it’s not pushed, its output just floats, but a pull-up on the output of the button will give it a defined “high” value so the input it’s connected to will read “high” and not pick up noise. When the button is pressed it becomes connected to a “low” voltage via a low-resistance path which will “win” (because its resistance to ground is lower than the pull-up resistance) and “pull” the output of the button low.

@StarCat As I mentioned, the definitions people use seem contradictory to me. Either, that's because I'm dumb, or, they are contradictory. The pull up resistor is usually (in schematic examples often used) just after the voltage supply, in series. So, your statement "a low-resistance path which will “win” (because its resistance to ground is lower than the pull-up resistance)" seems false, since path of least resistance is still through the "pull up", it just bypasses the output.

@Community I am extremely specific in the question. I also reference an example where a pretty famous electronics YouTuber mentions that it is exactly like I suggested. He mentions, and this seems true, that a "pull up" is actually strongest when it has zero resistance. The reason some resistance is used, is, like I wrote in my question, to avoid short circuiting. But the reason the "pull up" is used to start with, is, like @ dknguyen wrote, to avoid floating logic issue from noise and inductance and such. Feel free to claim that I am not specific, but I really am.

@Community I now rephrased the question as that "floating logic issue" part of why resistance is added to the voltage supply, is only half of the topic of "pull up resistor", from what I can see. And that my question is specifically about the other half of it, why resistance is added (the "floating logic issue" itself being solved by the default-on wiring rather than that it has some resistance. ) Maybe people get what I'm saying here. Peace.

@LearningBasicComputerScience the "path of least resistance" refers to when the button is pushed. The path to ground through the closed button will be much lower resistance (essentially zero). But I think this does not answer your question (which is not very clear to me at the moment). It's all basic electronics however.

@StarCat That statement still makes no sense, in most schematics used to explain "pull up resistor", since the current is forced to take the path through the "pull up resistor" regardless. That having a path of least resistance to ground lets current bypass output to some logic component, this is true even if you have no resistor at the voltage input. If you Google around, basic electronics stuff, you can see that the best "pull up resistor" actually has zero resistance, like I mentioned multiple times. But it causes issue of short circuiting in the "logic low" state.

Like I said, what you ask is not very clear to me. You seem to define "pull-up" differently to how it is normally used.

No. I don't. I point out that how "pull up" is usually defined, zero resistance "pull up" is the strongest one. What I ask is really clear, and, a logical question to ask for someone who just discovered the concept. The reason for "pull up" design pattern (inverting logic high and low relative input) is to avoid "floating logic", "disconnect", but, the reason for "pull up resistor" is to avoid short circuiting in the "logic low" state. This seems to be extremely basic electronics knowledge. electronics.stackexchange.com/questions/76376/…

The claim that a short circuit is the strongest pullup is TECHNICALLY correct but is so irrelevant when a pin is being driven that it is close to being rubbish. || A pullup is a device (usually a resistor) that ensures that the state of an undriven pin is "high". In a digital system it is seen as a logical high and in an analog system is seen as being high enough that it is indistinguishable from a driven high signal. || A pullup is a comprise between being so wishy washy that bias currents and noise etc negate its effect and being so 'stiff' that it adversely affects circuit operation.

@RussellMcMahon As someone learning the concept, it seems very relevant. If you look at my reasoning: "floating issue" is from "disconnect". This is resolved by reversing logic high and low voltage. I looked at countless content on that, it all says the same thing. Since this issue is resolved even without a resistor, it is just a matter of inverting the way switch changes voltage (inverting logic high and low voltage), it seems to me, 100%, without being irrelevant at all, that the "pull up" design, is distinct from wanting to insert a resistor. And, done for different reasons.

@RussellMcMahon I think people in electronics might miss how nonsensical the explanation of "pull up resistor" tends to be. You can see it repeated all the time that people have a hard time getting it. And, I'm not a dumb person, so, I usually get things when they're explained simply. That "A pullup is a device (usually a resistor) that ensures that the state of an undriven pin is "high", the point is, that is achieved just as well without the resistor. It is the inverted logic high/low circuit design that achieves that.

Saying that "the resistor does it" is like saying my glove is lifting the coffee thermos, when I'm out hiking and taking some coffee. It isn't my glove, it is my hand attached to my arm. The glove is doing something else (insulating against cold weather or warm coffee mug... ) The reasoning is wrong. The resistor does not force the logic high, the inverted circuit design does, also without resistor, actually, technically, even better without one.

If the best "pull up" to "force logic high" has no resistor then clearly it is not the resistor that forces "logic high" and explaining it like it is might confuse people since it is actually false. At least it was to me, and I think I get the concept now. So, anyone interested can take that as feedback from how one noob reasoned.

Also, I have never heard anyone say zero resistance is the best pullup. The strongest maybe, but not the best because it is useless.

Of the last 3 paragraphs, only your second last paragraph has potential issues. We say no current is consumed as an approximation. It means virtually no current flows into the input. If you sample blood, do you take half the person's blood? Or such a small amount that it is essentially zero and can be ignored? Same thing, The more current you take, the more the voltage gets distorted due to resistances.

@DKNguyen Thanks for the reply. That was the last thing I was wondering about. Having taken in at least some stuff now, I still think it is easier, for me that is new to all this, to understand a concept "pull up" as the logic inversion that removes a disconnected state with floating logic issues. And, the introduction of a resistor into that circuit, as a separate issue done for separate reasons, specifically to not waste electricity. I'm open to that I might be wrong on something, since I'm new to all this, but that's how it adds up to me at the moment.

@DKNguyen It's what most schematic examples of "pull up resistor" show. A switch that when closed is "logic high", is disconnected when switch is open, and causes "floating". So the high/low output has to be inverted, it is high when the switch is open, and low when it is closed. Still, thanks for the replies and helping me get a bit more.

It is best to think of it as low output when switch is closed and floating when it is open. Then the resistor pulls it high if it is floating. The switch+pullup resistor inverts, but the pullup on its own inverts nothing so if you talk about the pull up inverting no one will understand what you mean.

@DKNguyen Doesn't seem like that to me. All examples I've seen are solving the "disconnect" issue by making logic high the default when some switch is open (off) instead of logic low. An unintuitive design choice if not for the issue of floating. What "pulls it high" is that reversed circuitry. Not some resistor.

I actually have no idea what you are trying to say in that last comment. I don't mean that I understand your words but not your point. I literally mean I don't understand the combination of words you are using.

@DKNguyen usually if there is too much discussion here, it falls on the OP. These two both solve "floating" issue, imgur.com/a/rKqELlf. Because there is no "disconnect state". But one has no resistor. It will short circuit at logic low, burning electricity unnecessarily. Seems to be why a resistor is added.

In the real world, the input of a gate will have some parasitic capacitance. Thus, the pull-up resistor will charge this capacitance, to ensure we have a high state at the input when the switch is disconnected. Additional the gate can have an input leaking current that will flow from the Vdd via a pull-up resistor in the gate input.

@G36 What I'm trying to point out is that what "ensures we have a high state at the input when the switch is disconnected" is not the resistor, imgur.com/a/rKqELlf. The resistor is just added to make what actually ensures that high state work better and not short circuit when the switch is closed.

The job of this resistor is to "pull" the input "up" ( to Vdd). And this is why we call it a pull-up resistor. And we cannot use a wire because the circuit wouldn't work properly. The same is true for this case electronics.stackexchange.com/questions/586734/… Without T1, we could not have both states ( low and the high) at the output.

@G36 As I describe pretty okay in the question and my comments, it does not seem that way to me. What "pulls up" the input to Vdd, is that the switch is not placed in between Vdd and whatever logic component receives input. To avoid ambiguity and to not seem like I want to waste comment space here (I don't), I have illustrated that here, imgur.com/a/c9HX8W9. Figure 2 is what does what you say the resistor does. I add resistor in figure 3, where Vdd is already "pulled up" at figure 2 step.

But the circuit from fig 2 will not work. The fig 3 circuit will work because now we can low or high state at the input. Fig 1 will work with TTL gates.

What different "name" do you propose if you don't like the jargon term "pull-up"?

In summary, the role of a pull-up or pull-down is to ensure a valid logic level in the event that there is no active driver on a net. It is often important that the resistor bias can be overridden by drivers on the net such that the opposite logic level can be achieved. For this second reason, common pull-up values are relatively large resistors: 4.7K or 10K ohms, for example.

Why does fig 2 not work? Because when the switch is closed, Vin is connected directly to ground. The power supply providing Vin will be in current limit (best case) or on fire (not best case). As you probably know, if a power supply is not regulating properly, you cannot depend on anything working correctly in that circuit.