Gaaah! I knew you’d say that! Fine! It’s neither!
To be fair, I can’t blame you - even qualified engineers get this one wrong. In this specific example Mr.Mehdi will be playing his Oscar-winning role of “A Load”.
To explain what you’ve just watched (in case Electroboom’s explanation was insufficient), in theory the supply was supplying up to 150A of current, so in the absence of any resistance this is the amount of current that should flow. However, because the voltage was incredibly low, it was insufficient to overcome the resistance of his tongue (measured in Ohms), so the flow didn’t fry our favourite test subject. This ends today’s lesson in physics.
Last edited by Foxi4,
Gaaah! I knew you’d say that! Fine! It’s neither!
To be fair, I can’t blame you - even qualified engineers get this one wrong. In this specific example Mr.Mehdi will be playing his Oscar-winning role of “A Load”.
Haha that's a funny video. We're both right and I'm partly wrong too "in order to have high current you need high enough Voltage" or something similar was what he said. So for more current you do need more Voltage. Resistance indeed has to do with it. (a lot)
The easiest way to explain it is that in a realistic setting you can’t have one without the other, and we haven’t even touched upon voltage drop yet. It’s a fine science.
Voltage "delivers" the current. No current means nothing gets "delivered" and no Voltage means the current can't go anywhere. I think that's a good explanation for that. Nonetheless the OP still thinks heat is the only thing that damages it. They should watch this video too. There's nearly no heat involved in that example.
Well, yes and no. The OP isn’t *entirely* wrong when saying that it’s the heat that destroys the component - when you pass current through a component, you generate (waste) heat. Pass excessive current through any resistor and I guarantee you that it will light up like a lightbulb before it pops, if momentarily. If you were capable of instantly removing that heat with magic, the component would not pop despite running out of spec - the problem here is that we’re not magicians. A heatsink won’t remove all heat from the entirety of the component instantaneously, there’s a junction between the component and the surface. Even a discrete sensor won’t tell you exactly what’s going on inside a given component - they measure temperature at a specific junction. This is why any datasheet normally comes with more specific parameters in order to provide a ballpark figure of the difference in temperature between the package and the actual chip inside of it (if such information is relevant). That’s splitting hairs though, I think this point was already made. There are some other considerations, like diffusion of the dopant material or even internal shorting - electricity can produce arcs, after all. Lots to talk about, nothing relevant to this thread.
More physics, not that it really matters in a thread like this:
https://en.wikipedia.org/wiki/Junction_temperature
Yes. Heat accelerates degradation but without a direction. Current is what kills. You got 3 things to consider: heat, current/Voltage and speed/frequency. Reduce one and you can increase one with the same rate. Go sub zero and you might be fine running stress tests (high current) with insane Voltages like 1.8, or even 2V. (der8auer with Intel Skylake).
Honestly, we’re not even talking about chip degradation, we’re talking about spontaneous failure. To circle back to what we talked about before, heat isn’t so much a “consideration” as it is a direct result of what you’re doing - if your circuit is pulling 40W, that 40W has to go somewhere. Prominent figures in the OC scene like der8auer know this, otherwise they wouldn’t be running their components at insane values.
The SOC is actively cooled, the rest of the circuit ostensibly isn’t, which is the problem here, and not something you can address in software. Not sure what we’re discussing here anymore considering risk management is up to the user.
The heat is a result from the current. That has to go somewhere. It leaves the chip in the form of heat. Well, the heatsink is taking it away, it's not like it's going away on its own (without heatsink it's the air that takes the heat away). This is far more extreme than any Switch would see, though. It can't even deliver more than 1.5V. Main facts remain license, GPU not doing anything but more power being drawn and heat, and still a lack of warnings. Saying you're or the mods aren't responsible isn't really warning for what could happen, like spontaneous failure or degradation. Also, the traces on the motherboard are only designed for 18W. 40W would damage them too. Der8auer and all those aren't planning on keeping them like this, only short testing. That doesn't matter as much as it does now.
The only issue is licensing. Nobody is responsible for how a piece of software is used besides the end-user when the end-user is informed that the software can damage their system. The burden of conducting necessary research is, and always has been, on the end-user. If someone foolishly runs their Switch full blast regardless, I’m afraid that’s their problem. The warning was provided, and was additionally elaborated upon by another mod by editing the OP.
Ultimately I think the only correct resolution to this, in my opinion unnecessary, kerfuffle would be to provide links to relevant repositories so that the end-user downloads each individual component on their own, as opposed to providing a zip package - this resolves our GPL issue, although it will require a more extensive installation tutorial. @Cooler3D should follow that approach in order to avoid further complaints, and I’ll give him ample time to edit the post in this manner.
In my estimation this solution makes everyone happy-clappy-chappy. If @Cooler3D doesn’t respond in a manner that I find satisfactory, I’ll remove the package, but as my good uncle once said, “haste is only advised when trying to catch a flea”.
KazushiMe allows the redistribution of his package, if it complies with the GPLv2 license and has proper warnings. He already has the complete package and this is almost identical, only not updated to include the latest improvements KazushiMe made. It's not an issue if the OP would do would you said, though I don't see any point in it. Uploading the source code to Github and link the releases page here is enough (although KazushiMe does the exact same thing so that's a bit pointless too). Either way, complying with all licenses (which includes proper warnings) would solve all issues. Risks are then truly for the end-user.
I’m speaking purely from the perspective of uploading the content *here*, Git has its own requirements which don’t concern me. If the OP would prefer to keep the whole thing on GitHub, they also have that option.
That’s correct, which is why it would be significantly easier (and faster) to link to relevant repositories instead. It provides clarity and compliance.
That is what I did say, chips use more amps when overclocked
Voltage control is for undervolting, which I also said.
Basic electronics, the voltage is set by the things providing power, the amount of amps drawn is set by the thing using the power (up to the limit of the thing providing the power). If you draw too many amps then the voltage can drop, but that isn't really a way of controlling the voltage. Certainly on a switch you wouldn't want to then try to increase the voltage to counteract that.
Last edited by smf,
I’m not convinced you know a whole lot about “basic electronics” based on what you’ve said, not that it matters. You seem to be under the impression that overclocking is achieved by some kind of current control, that it’s actively “set” somehow (at least that’s what it seems like), which is not at all how it works - an overclocked chip can possibly (not necessarily, manufacturers often leave a lot of headroom on the table) require more current, but only by proxy. I’ll post a more detailed explanation on how power delivery works, without getting too much into the engineering side of things to keep it understandable for any future visitors to this thread.
The “amount of amps” isn’t “set” by anything in a typical CPU/GPU/SOC arrangement - power delivery is controlled entirely via voltage regulation because that’s the simplest way to achieve the desired result, and this translates to nearly all of the consumer electronics you could think of, including the Switch. You can’t “increase the current” without increasing the voltage - a conductor will provide as much flow as its resistance permits it, that’s just physics. The actual, real life current flow is load-dependent, so either there is a load or there isn’t. In peak demand conditions, when the circuit draws as much as it possibly can, in order to pass more current you *must* necessarily increase the voltage, and consumer electronics do that automatically without your input, within parameters specified by the manufacturer. Conversely, if a chip is idling and very little current flows you can afford to drop the voltage, and consumer electronics do that as well. The chip’s consumption directly correlates with the load, but it is not actively “controlled”, voltage is. A processor is just a bunch of various transistors formed into logic gates and etched in silicon. Either they’re actively opening/closing, and consuming electricity to do so, or they’re sitting idle and there’s very little load, so the current flow drops, *regardless* of operating voltage. Current cannot flow within a circuit if there’s nothing in it that actively demands it. Anyone can test this by hooking up any electronic device to a Kill-o-Watt and observe the power consumption in relation to the workload the device is performing.
In practice, power delivery is achieved with a voltage regulating module, or VRM. A basic VRM consists of a MOSFET (let’s keep it simple and call it a transistor, an electronic on/off switch), an inductor to stabilise current (it’s a coil of wire, usually encased in a ceramic package), a capacitor (nowadays solid state, but you see the occasional electrolytic still) to stabilise voltage and some kind of voltage regulating controller. The first component will dictate the voltage in the circuit by rapidly switching, the next two provide smoothing, because computer chips don’t like current or voltage spikes, the last uses built-in logic to calculate the frequency of switching required to operate, as requested by the processor in question. You have two active components, the transistor and the controller, and two passive ones, an inductor and a capacitor. By changing how quickly the transistor switches on and off, the circuit can output higher or lower voltage, and thus supply more or less power to the chip, based on demand. In layman’s terms, if you supply a VRM with an input voltage of 5V, but the frequency of switching is such that the VRM is only in the “on” state 1/5th of the time, the output of that VRM will equal 1V, and the maximum current will depend on the resistance of the circuit - I = V/R, current equals voltage divided by resistance.
As far as current control goes, the only thing you can usually observe in a setup like this is overcurrent protection. It effectively functions as a fuse which trips when current grossly exceeds the spec, but that’s usually located closer to the input, not the VRM itself. The reason for that is that it’s kind of irrelevant *where* your overcurrent protection is - draw is draw. Meanwhile, for the purposes of maintaining precision (avoiding noise, voltage drop etc.), the VRM is located closely to the chip. It’s in the name - Voltage Regulator Module. It’s the voltage that is actively controlled, higher or lower headroom (not flow) for current is the consequence of regulating said voltage. Voltage control is not just for “undervolting” - that’s one possible use case. Often times you can find that a chip will operate exactly the same at a lower voltage, so there’s no point in supplying one that’s higher than needed, it’s a matter of fine-tuning. More importantly, voltage control is there for the circuit to operate at all - without it you would have a static voltage which is incredibly wasteful and not optimal, since as I’ve mentioned before, you’re aiming at the lowest possible voltage required to operate at any given moment.
The reason for this setup is simple, and it’s remained unchanged for god knows how long. A typical CPU/GPU/SOC works within very tight parameters in regards to voltage, usually somewhere between 1-2V, closer to 1V. Even a small change of a mV can drastically change how the circuit operates because we’re talking about billions of transistors on a very small silicon die, it’s all highly sensitive (hence smoothing). That’s obviously not the supply voltage, which is much higher, so you need some kind of circuit to take whatever supply you’ve got and buck it down to a desired level, with extreme precision, and with the ability to adjust dynamically. The VRM does just that. By supplying a higher voltage, you can overcome resistance easier, and as a consequence of that, more current *can* flow and more power can be delivered, but that doesn’t necessarily mean that it will - it depends on whether it’s needed. In this regard, the Switch is no different. Its power delivery is, fundamentally, built in the same manner as what you’d see on a desktop computer, or any other similar consumer device.
Long post, so tl;dr - an overclocked chip will likely have a higher power consumption than a chip running at stock frequency when at peak - it’s doing more work, or the same amount of work in a smaller amount of time. This may or may not equal higher current - that depends on the voltage it’s running at and the workload it’s performing. 15A at 1V equals 15W, however 12.5A at 1.2V *still* equals 15W. Higher voltage, lower current, still the exact same amount of power, so clearly, this is a matter of how much is the demand versus how much can be supplied more so than anything else, aforementioned waste aside. It is likely that overall consumption will be higher, but that’s not a statement on current, it’s a statement on power. A blanket statement like “it consumes more amps” is misleading, amperage is not a measurement of power, it’s a measurement of current. One can’t make a statement like that without explaining the circumstances or understanding the circuit, because it’s not always true.
You've obviously not got any understanding of anything about electronics.
Because "it consumes more amps" is exactly what happens when you clock faster. A circuit can't just consume more volts by increasing the load.
You can vary the voltage to deal with stability issues and to reduce heat (if you undervolt). But just increasing a clock speed by itself won't do that. Under volting would have most use, because the problem for the switch is the heat and watts (unless you plan on watercooling it). Undervolting is what most people do these days with PC's.
I suggest you stop now. Your wall of text suggests you're desperate to be correct, but you've failed.
Last edited by smf,
Wow. Okay, “consuming amps” or “consuming volts” is not a thing, an ampere and a volt are units of measurement. The reason why you’d raise voltage when overclocking at all is specifically because the circuit doesn’t have sufficient current carrying capacity, so you can supply more power at a higher voltage, even at the same current, if needed. Please, continue not understanding the subject, I’ll add physics and electronics to the list of things you’re not good at.
Trolling as per usual.
I just don't have time to pull apart all your incoherent babble & nobody else can too.
My point is that overclocking causes the gates to switch more often which requires more amps. The volts are set by what is producing the power. You have a hate boner for me, so you always try to contradict me. But if you have a psu providing 5v you can't suck 6v through because of an increased load.
You literally cannot disagree with this as it's basic physics. A simple google search can confirm it.
Your wall of text is just pointless for everyone. I'm sick of you doing this ALL THE TIME and I really wish you would grow up or disappear, the fact you are a mod is just a sick joke.
Last edited by smf,
This all doesn't belong in this thread anymore. Where did Foxi4 say the Switch draws Voltage? Voltage is supplied by the VRM's and it draws the current. It can, however, ask the VRM to supply a certain amount of Voltage. This is usually called the VID. That's not the point here, though, and neither is it mentioned anywhere.
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