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Hmmm
Hmmm
00:19
The power source is just a 12v battery connected to the D36V50F5 to step down the voltage to 5v. If I replace D36V50F5 with LM2596 it works.
the only tool i have is a cheap multimeter
W5VO
W5VO
01:18
@Hmmm And what does the cheap multimeter say the output voltage of the DV.... module is?
Hmmm
Hmmm
5.1v i tested another D36V50F5 to make sure i don't have a faulty one
the lidar works with it but sends garbage data
it communicates via UART
W5VO
W5VO
01:31
random data and not just noisy data? No idea. I'd probably check the power-good pin (give it a pull-up) and see if the converter thinks everything is OK, but you'll need something faster than a cheap DMM.
 
3 hours later…
Scientist Smith YT
Scientist Smith YT
04:03
@Shalvenay Ok thanks :)
@Shalvenay & @W5VO & @ThePhoton & Anyone else... I want all hands on deck for my next question because it won't be easy to answer.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: I want to make a single transformer welding machine using a Microwave Oven Transformer. The transformer will be a 1,800 watt transformer. It will run on 125 volts using 20 amps.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: My parameters are as follows: Voltage Output Needed: 50 - 60 volts AC. Current Output Needed 230 - 265 amps. I will be using 10 AWG or 12 AWG THHN Stranded copper building wire to wind the secondary coil.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: The secondary has a height of 2.5 - 3 inches roughly with 1.5 cm width roughly. Maybe close to 2 cm. And a length of 2 inches roughly.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: I do not know how many windings are on the primary and I can't calculate or estimate it close enough. The current limiting shunt will be removed from the transformer to allow easy saturation of the transformer core. So I can accomplish a much higher output current.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: Operating frequency: 60 Hz. The resistance of the primary is indecisive. So I don't have a measurement for that that is accurate enough to say for sure.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: I know it is possible to get 300 amps from a single transformer using 12 AWG and 10 AWG wire. But I'm not sure how many windings I would need for the secondary to get 50 - 60 volts and the 265 amps I'm shooting for.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: Especially with the Information I know, I'm very uncertain. I tried using the formulas for calculating output current and voltage as well as turn ratios. But I'm getting numbers that don't line up and won't work, or don't give me what I'm needing. And the numbers it's giving me for the voltage and current I know and feel aren't accurate to what I will actually get.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: I know there is a way to figure it out from what I have. But I don't know what that way is. (Other than going through testing. But I don't want to waste my time doing that because it'll be a lot of testing) Any help or suggestions that are solutions?
The Photon
The Photon
04:33
@ScientistSmithYT, if your input is 125 V / 20 A, you're not going to get more than 40 A out at 60 V.
You seem to think that conservation of energy is a rule that only applies to other people, but I am pretty sure it applies to you too.
Scientist Smith YT
Scientist Smith YT
@ThePhoton Well it applies. But I was able to get 570 amps out at 42.6 volts with 4 transformers like that. So how did I do that?
The Photon
The Photon
@ScientistSmithYT I dunno. You must have changed the laws of physics in your lab.
Scientist Smith YT
Scientist Smith YT
@ThePhoton 40 amps is what the calculator said when I did it. But that doesn't seem right. I couldn't have changed the laws of physics. So there has to be an explanation. And I can't find that darn explanation or reason why it does work.
@ThePhoton I've looked and looked and searched and searched. I've asked and asked and read and read. I've tested and tested. I've experimented with theories and theories. What else do I try besides giving up?
@ThePhoton There has to be an answer. There just has to be.
@ThePhoton The formula for my larger welding machine said my maximum output current was 198 amps and it would use 65 amps. But It actually uses 40 amps and outputs 570 amp. Which is way off. My guess is that taking out the shunts makes the transformer output more power by utilizing more of it's energy it transfers in the core. That's my only explanation.
Which is possible. But that wouldn't explain a secondary part. Which is the power usage and voltage output
So I'm stumped there.
Like I've been trying and trying to figure this out and find the reason or explanation as to how and why it does this for a couple years now. And there's road block after road block.
My other idea is that the formula takes into account the least amount of energy the transformer can transfer and output. That would make sense. But then why have the formula if it calculates the lowest possible output factors? Wouldn't we want to see how far we could go for the highest amount. This question definitely isn't an easy one, I can say that for sure.
@Shalvenay & @W5VO & @ThePhoton & Anyone else: If any of you guys have any ideas or theories on this please share them with me because I could use a load of new ideas and theories on this. :)
@ThePhoton Thank you trying to help. :) I need some rest. I've probably been trying to find a solution for too long.
 
9 hours later…
W5VO
W5VO
13:47
Occam's razor says you're not measuring high current correctly. Or voltage. Or saturating your transformers is giving you nonlinear waveforms that can't be diagnosed with a DMM. Or power factor.
W5VO
W5VO
13:57
@ScientistSmithYT And don't spam notices like that. It's annoying.
Hmmm
Hmmm
14:40
what does NOPB mean? there is LM2596SX-3.3 and LM2596SX-3.3/NOPB D
 
2 hours later…
W5VO
W5VO
17:04
@Hmmm Lead Free (No Pb)
It's unlikely that on a recently purchased part there will be any difference (both will be lead-free)
Scientist Smith YT
Scientist Smith YT
17:26
@W5VO What do you mean about Occam's Razor? And how am I not measuring current or voltage correctly? How do I measure the current and voltage correctly? I measure the voltage in a sine wave because it is a sine wave. Are you saying that there is no solution to this problem? And I got it, I won't give too many notices.
 
6 hours later…
W5VO
W5VO
23:11
@ScientistSmithYT Here's my assessment of the situation: You need someone to come in, see what you're doing, and probably bring their own measurement equipment. You have big enough gaps in your knowledge and/or equipment that I'm all but certain you're not going to solve this with help from a chat room.
Occam's razor
Occam's razor, Ockham's razor, Ocham's razor (Latin: novacula Occami) or law of parsimony (Latin: lex parsimoniae) is the problem-solving principle that "entities should not be multiplied without necessity." The idea is attributed to English Franciscan friar William of Ockham (c. 1287–1347), a scholastic philosopher and theologian who used a preference for simplicity to defend the idea of divine miracles. It is variously paraphrased by statements like "the simplest solution is most likely the right one". This philosophical razor advocates that when presented with competing hypotheses about the...
You have presented us with a circumstance: "I made these measurements, and I have created power." Now either you have broken a law of physics that people have been trying to break for millenia, or you made a mistake somewhere.

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