@NickAlexeev Huh!? What!? Hmmmm....??? Why was it called "serial box"? I only read the first part of the article so far. But I'm guessing you might know why it was called that.
@ScientistSmithYT Possibly an error in transcription. Another person in the linked interview refers to the same stage in development and calls it "serial bus." Or maybe they really did call it "serial box." The interview mentions that many of the early meetings were "breakfast meetings."
@ScientistSmithYT USB - read that then follow the links and look up the referenced standards.
@ScientistSmithYT About your transformers: Assuming you have on transformer with 108 turns primary and 4 turns secondary, then you could expect to get around 540A out of the secondary. 108/4=27. Multiplying 20A in by 27 gives you 540A out. But, only 4.4V out.
You have to measure the voltage under load, not when the circuit is open.
@ScientistSmithYT You missed the book: janaxelson.com/usbc.htm . It's a 524 page book that simplifies the USB protocol to the point you can use it to design electronics.
The USB 3.2 specification itself is 548 pages, but there's a ton of additional documents regarding power delivery, testing, and compliance.
@VoltageSpike If you want high power, you probably have to exchange some messages with the host to get it.
And every cable connected to your device, whether power or data, is a possible source of conducted emissions, or antenna for radiated emissions, so it definitely affects your FCC tests.
@VoltageSpike Anyway, there's no government regulation of USB itself.
If you want to put a USB logo on your device, you need approval from the trademark holder (USB IF, AFAIK) and (again AFAIK) you'd need to do compliance tests to get it.
So I've got a function describing voltage across a capacitor. It's u(t) = e^(-20t) for t > 0. What confuses me about this function is that voltage across a capacitor is supposed to climb over time. Yet this function drops to near zero quickly
Anyways, since I'm asked to find i(t), the current function. Using U = Q/C, I rewrote charge (Q) as I dt (integral of current) and got: U = 1/C * (I dt). Then I assume that d/dt (U) = d/dt (1/C * I dt), and so d/dt (U) must be the same as u(t).
@W5VO Ah yeah your'e right there actually. I didn't remember that
That finally gives me: [ e^(-20t) ] = 1/C * I. So I = C * e^(-20t). The only thing I'm unsure about is whether I'm right in stating that the function for current at time t is then d/dt I, so I have i(t) = d/dt (C * e^(-20t)).
Reason being is that I get a negative function for current. That doesn't make sense to me because if voltage is decreasing over time, then current should be increasing no?
Well actually its an 'increasing' function with a negative coefficient. But it approaches zero.
@Micrified Positive capacitor current, or current flowing into the capacitor, charges it up. Negative capacitor current represents a discharging capacitor (your case)
@Micrified Useful for future context, that is basically a resistor-based discharge of a charged capacitor. See en.wikipedia.org/wiki/RC_time_constant for more examples and formulas
If the goal is Laplace/equation solving though, then sometimes the actual use will result in logical problem solving steps in courses that are not wanted.
Well I am only given the voltage at time t function: e^(-20t). If I integrate this to get voltage overall I get a negative value because the negative sign is brought out,
Wait so if I'm computing R = V / I and my I is negative because my capacitor is discharging. Should I flip the sign for the current because I'm compute it as going into the resistor positive terminal ?
you might draw an intermediate step where you label the loop current i1, draw the loop current direction (e.g. clockwise or counter-clockwise), and then you'll get a sign flip that way as well
It's a good strategy for showing your work. If I were solving something like this, I would probably adjust the signs if I got the wrong answer. As a student (I'm assuming you are one), you must show how the signs work out for the right answer.
@JRE The input according to the formula and the usage values exceeds the very maximum input value of 2,460 watts. The breaker trips at 2,400 watts. The whole device uses >3,000 watts on input according to the formula and input values. Which is impossible, because it would trip the breaker.
@JRE I take the output and input values of voltage and current input and output, plug them into the formulas. The input is wrong, ot it has to be. And now the output I'm doubting, but the output seems close enough. But if the formula said a crazy input value my guess is that the output might be wrong.
@JRE These transformers are not effected enough by voltage peaks as other appliances are. So the voltage peak doesnt help as much to deliver more power.
@JRE One thing to take note of is when I turn both of the transformers on at the same time, it draws up to 31.5 amps nominal. Its max could be up to 35 amps.
@JRE I think you're seeing what my big problem is now. These numbers from the formula are not lining up with the input and output values I am getting. At first I thought the core was a problem, it wasn't, then I thought excess fields from the high amp output was a problem and was interfering. But the gauss readings were normal for what should be expected. So I went to maybe the coil was a problem, it wasn't. Then I thought many other things were a problem. They were also not a problem.
@JRE I asked my electrical engineering buddy, he doesn't know. But he is in school for it now for his masters. So he's still learning. I've asked electricians. They have no clue. I'm so determined to figure this out, but I have no clue now as to where to start back up at. I've started, and I've come to an end seemingly. It's frustrating.
The sole purpose of a circuit breaker is to save the wiring of the facility in which it is installed. It's made with the precision required for that task - (low). Any time it keeps you from destroying your device is bonus.
@ScientistSmithYT Honestly, I don't understand the problem. You exclaim loudly that it isn't working right, that the numbers make no sense, but you don't show your setup, your calculations, or your results in a comprehensible way.
"When you short circuit the secondary of a MOT, the primary presents essentially the leakage inductance. This will limit the current drawn to 2 or 3 times the normal operating current, the right ballpark for 'fun with MOTs', and generally low enough to leave fuses/breakers intact. "
That is an effect of saturation in the core of the MOT.
Why might it be that an LED connected to a GPIO pin (with a 330 Ohm resistor in series with it to ground) doesn't light up unless I press the probes of my multimeter against ground and the pin that is supposed to be lit up?
Measure the resistance from the ground pin (on the dev board) to the LED (negative terminal), and measure the resistance from the I/O pin (on the dev board) to the LED (positive terminal)
As well as the obvious: you're sure you have the LED turned the right way around?
@JRE I'll take some pictures of the set up I have and A diagram of the wiring as well as a picture of the wiring itself. I'll show you everything again. I say again, because I've shown it a long time ago on here before. But I'll send them again.
@W5VO A general breaker isn't calibrated persay. But ours are. We didnt just get some random crap breaker at the store.
@JRE I don't want my images that I send to be thought of as spam. Because I'll need to send more than 3 images and explanations for it. I was wondering if you could create a seperate room for this? So those who don't want to look at all of it dont have to and have a choice.
@JRE It's not super simple for my project to just use 1 image. It wont help me or anyone else help me solve my problem. I want you and whoever else who wants to help know exactly what I have. I don't see that 1 or even 3 images will cover it in its entirety.
I'm not sure if I have permission to create a room, so that's why I'm asking you.
@W5VO Split phase in the U.S is sometimes used on purpose. But that's just an fyi, not very important. It is used in some cases though. In those cases it is a good thing. Some in line transformers on powerlines use a split phase method in some areas because of available power overall. I see it in Ephraim Utah everywhere. Down there they don't have as much power availability as we do in the city or outskirts of the city.
