Nice for a video down the track if you could build up those other two solutions (TI and OnSemi iirc?) and then measure the three different options. Four if you count the full bridge.
Turns out all that funny business is not coming from the product consumption, and with a moments thought that is obvious, more investigation required, likely a proby thing.
A pure R-C-Zener circuit should not create that mid frequency stuff, just the front and back porches.
The zero voltage crossings look alright. The mid frequency on the top is a bit odd. Nevertheless, the waveform will be non-sinusoidal. Grab the mid frequency with the cursors. Is it an exact multiple of the line frequency? Could very well be a quirk with the probes and this particular measurement.
But it isn't power. Please think back about how a generator works, the equations involved (induced voltage due to moving a wire through a magnetic field) and go from there. Your numbers don't add up and it is never ever going to be anywhere near 100MW (not even including I2R losses). Just do the math. For starters let your scope multiply the V and I trace to calculate the power and then calculate the RMS power from that. That is the real power and that is what the mains needs to supply.Not necessarily true. That is the only power the mains needs to supply at your house maybe. But the I^2R losses in the grid are a result of the current not real power.
Not necessarily true. That is the only power the mains needs to supply at your house maybe. But the I^2R losses in the grid are a result of the current not real power.
I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.
I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.
I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.
Think about it again, and how I did the calcs in the video for a home, and then get back to me.
I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.
Think about it again, and how I did the calcs in the video for a home, and then get back to me.Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal. So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor. My simple, off the cuff, calculation already shows that the I2R losses for the smoke alarm are neglectible even if my resulting number is 10 times too low.
I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.
Think about it again, and how I did the calcs in the video for a home, and then get back to me.Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal.
So please enlighten us with a real calculation including a good estimate of the actual grid losses. Hint: if small devices with low PF are deemed a problem then they would be subject to regulations regarding power factor.
My simple, off the cuff, calculation already shows that the I2R losses for the smoke alarm are neglectible even if my resulting number is 10 times too low.
Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?
Again, I'm done on this 100MW thing, I will ignore all further questions on it. If you want to talk harmonic PF, filtering, and low power design, please do so.
Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?Obviously not. But, once more ignoring I2R, you won't be burning a Joule of fuel more anywhere.
Which means that the claim that it "kills the environment" is a hyperbole at best, pure clickbait at worst.
The content between, say, 8:00 and 11:00 or so is confusing, misleading and sometimes just wrong.
I am not going to waste my time debating the semantics of grid impedance values, which is only part of the issue, knock yourself out. Your numbers aren't even close to being right BTW, even with your own narrow framing, I'll let you figure out why. Hint, current squared.0.5 * 0.08 * 0.08 = 0.0032W. Not entirely sure where he's wrong.
Think about it again, and how I did the calcs in the video for a home, and then get back to me.Sorry for my critical thingking but the calculations shown in the text of the video make no sense to me because you keep mixing Watts and VA as if they are equal.They are not equal. BUT if a product takes 80mA in measured RMS current, and that is NOT compensated for in the grid by filtering and/or phase correction, then to maintain the same mains voltage that power MUST ultimately be generated at some point, there is no free lunch there.
Thought experiment for you: Assume your grid is at almost maximum capacity (minus the 1.2MW real device power (1.2W * 1,000,000 devices)), and there is no more harmonic filtering or phase correction left to give, and you suddenly connect a million of these devices to the grid, where do you think the almost 100MVA magically comes from to maintain that 240V at the home? The current fairy?
QuoteThe content between, say, 8:00 and 11:00 or so is confusing, misleading and sometimes just wrong.I admit it was poorly explained.
But Dave, you waste more energy than even a million smoke detectors.
...
...
I watched it again, and yeah, it's just too confusing I think. I have removed that part with the edit tool, I hope it works. Will take many hours to process.
Is there a DC offset? The scope is AC coupled yet the trace is up.
The scope is DC coupled.
Although I am a bit perplexed why the negative half current is the same as the positive half current. But kinda makes sense in that the zener is just a diode in the half cycle, and sensor circuit is doing whatever
all the time to generate that crap.
BTW, the other Quell brand is near identical.
You can't ignore I2R at that point, that's the whole point
Exactly, reactive power does not cost you money, but it fully contributes to grid losses (unless it happens to be compensated by other inductive loads nearby). It is still not clear, though, how much real power the device consumes. Given that other devices can last 10 years with a battery I find 1W still pretty high - that would be 2.60 Euro/year, assuming the electricity price in my country.
It takes 5W to charge a 1µF capacitor 100 times a second.
100*1e-6*((240*sqrt(2)-15)^2)/2 ~= 5.2 joules/s
Where are the other 15W up to 20 VA dissipated then, in 100R and the zeners? 15W are many watts, they should be quite toasty. No?
@79.3mA and 1.36W total measured:
100R = 0.63W
So that leaves 0.73W for the zeners.
Ignore the load which is naff all.
Maybe a smidge in the cap ESR.
The rest will be copper losses down the system unless compensated for.
Don't think I'll go that far, it's obvious what results a HV regulator will give in terms of current vs the zener solution, no need to build and measure to prove that.