• Why doesn't wind reach 1038mph?

    esot.eric10/09/2019 at 15:50 0 comments

    weird thought:

    A) wind is caused by hot-air meeting cool air. Presumably, the less-dense hot air pushes the denser cooler air.

    B) the sun reaches the surface of the earth each morning, at a rate of 1038mi/hr

    C) the night-air is cooler than that which is heated by the sun

    Then: there should be a constant east-west wind following the sun's reaching the planet's surface, at 1038mph!

    Flat Earthers Unite!

    (Arising from thoughts on why pre-sunrise the sky was clear, but during sunrise we were engulfed in fog)

  • dummy-load power recovery 3

    esot.eric09/23/2019 at 21:25 0 comments

    decided to try paralleling the load-recovery "source" voltage with the batteries. There's a bit to consider. 

    For one thing, we've got a flyback inductor releasing power back to the input. At a 0.6A@5V dummy-load, that's 3W... But, of course, the, say 90%-efficient, converter needing loading is driven from a 40V battery-source, and thus the current going into that converter is much less than going out. 3.3W, 40V in, 3W, 5V out, so, roughly 0.6/8, less than 100mA. 

    So, now, we charge up an inductor from its 5V output at, say, 1A [because we need a 0.6A average load, but some of the time will be spent with the inductor discharging]. Now, when discharging the inductor we need to dump 1A as quickly as possible. But it's discharging back into the converter's input, which is far less than 1A, more like 100mA. 

    So, the thought, then, is to dump the majority of that current into charging a capacitor, then since the converter's input-range is 36-75V, that capacitor can charge to 75V. But, of course, this isn't free-energy, so once the capacitor's charged-up, it will begin to discharge, during that whole process dumping roughly 2.4W [assuming my circuit is 80% efficient] into the converter. But the converter needs 3.3W, so the inductor/capacitor obviously can't power the converter continuously. Once the cap discharges to 40V, the batteries take over.

    Meanwhile, the inductor's being charged back up to 1A, and the cycle will repeat.

    Now, all those numbers are rough guesses, and I've never designed something like this before, so it's coming on time to simulate. [And I don't have a clue how to calculate the necessary inductor/capacitor requirements] Like... how do I assure the cap doesn't exceed 75V, but that all our input power also goes out? Or, worse, how can this whole thing be properly-timed such that our inductor-charging and discharging times work, as well as the capacitor discharge time?

    so, I thought an easy test would begin with falstad's simulator, best thing I could find for simulating the 3.3W converter input is an ideal voltage-controlled current source. P=VI, right? Nah, I=P/V, and it *really* wants to try V=0. So, now I've a bit to learn about the tools.

    Ultimately, the point of the project, which led to this tangent, is to have a reliable and long-term power-source for my computer, among other things, so continuing down this tangent is a bit chicken/egg. That simulation's going to take some time to setup, meaning some time on the computer, which needs power. The 0.6A minimum load, realistically, is probably not a necessary concern to get this thing running.

  • Minimum load recovery pt.2

    esot.eric09/19/2019 at 21:48 0 comments

    Don't put the load-recovery circuit in series with the input.

    Quick estimates, assuming the DC converter's 90% efficient and the load-recovery converter is 80% efficient look pretty great. We can turn our 0.6A@5V=3W minimum load into only 0.93W load on the batteries!

    But putting the load-recovery circuit in series with the batteries... well, here's a quick thought to get er goin': 80% of 3W is 2.4W. And a 90% efficient DC converter will draw 3.33W to make that 3W [to then make that 2.4W]. 

    So now we have a 2.4W source in series with our 40V's worth of batteries... and the sum total is 3.33W, meaning the batteries are only outputting 0.93W! Great! 

    But P=VI, and since the sources are in series the current through our 40V batteries [at 0.93W] is the same as the current through our load-recovery voltage-source [at 2.4W]. Again, P=VI, I is constant for both sources, P is comparatively huge from our load-recovery voltage-source, so then V of that voltage-source must also be comparatively huge... to the tune of 103V, on top of the 40V. These DC converters have a wide input range, but nowhere near *that* wide.

    So, parallel's probably the way to go, but it seems a bit risky to directly-parallel DC-converter outputs [again, the battery packs are 3.7V with a 5V boost converter]. And then it seems a bit goofy to insert e.g. diodes or resistors... they'll add to the losses we're trying to recover [and likely still, even when the real load is on and the dummy load disabled]. But, may still be worth considering.


    And, it's kinda an interesting circuit I've come up with to do this... essentially a boost-converter whose feedback node is on the opposite side, measuring/maintaining input *current* rather'n output voltage [or current]. For my first attempt at a homebrew DC-converter, this'd be a doozy!

  • Minimum load recovery

    esot.eric09/15/2019 at 05:53 2 comments

    Thoughts on a switching dummy load, for a DC-converter, which redirects power back to the input.

    At a minimum load of 0.6A, we're talking 20.4W just to power the thing with no useful load attached!

    These are floating-output converters with a wide input voltage range, which, I'm pretty sure, means, if done right, the outputs can actually be placed in series with the input.

    No, we're not talking free-energy, here... we're talking recovery of otherwise burnt energy.

    Though, I've also pondered this same output-in-series-with-input to maintain the minimum input voltage while, say, half the series batteries are removed from the circuit for charging.

    Weird world.

    Oh, and this dummy-load would also decrease as the actual load increases.


    And then *really* vague thoughts on using a similar technique, and the scaling necessary, to replace resistors in many general cases. Please send me some moolah and some credit if you steal this idea!

  • So then I was like...

    esot.eric09/01/2019 at 09:12 2 comments

    "I should put a 12V lighter-outlet in this thing..."

    Waitaminute.... one of the main points is that there are some things I don't want on a randomly-self-ejecting nor easily-bumped power plug.

    [Of course, that'd've been easily resolved with in-dash banana jacks, which somehow hadn't occurred to me until weeks into this project]

    And, of course this bright idea came from... "how should I connect [the thing that needs a reliable power connection]?"

    But I think I have come to some pretty decent design-plans, many of which differ dramatically from my original goals.

    E.G. I was planning to have two 5V 6A outputs, one for running my itty bitty computer, another for [USB, mostly] peripherals. But now I'm thinking of one 12V 6A run to the computer and two 5V6A dc-converters inside. Then I'll also have 12V available for peripherals, and only one long cable. And the 12V source has sense-lines, so 12ft later 12V should still be pretty clean. 

    Now debating cabling... am kinda low-budget and mostly trying to work with things nearby... Am thinking ethernet cable with pairs combined at each end for power, and sense lines on another twisted pair. Debating the fourth pair. And something about twisted-power-pairs [nevermind three pairs in parallel tied together at the ends] doesn't seem right to me... inductance? And large, plausibly somewhat largely-varying, currents in opposite directions through a 12ft transformer?

    Meh... even though the AWG/Current tables are "very very conservative", a 0.5A per wire rating is nowhere near the 2A that'd be on each, if I used 3... so, some reality-check... 2.5ohms/100ft is 0.25ohms at 10ft [doubled, current out and return]... if I used a single wire for each, that'd be a 3V drop! So, even with three wires, each, still a volt. I guess not so bad, but still those other concerns. [And where's 12W of heat go? And, yeah, most ethernet cable is pretty stiff...]. But 4 beefy wires seems goofy, where yah gonna find two beefy and two thin in a bundle?


    Then again, use beefy-enough wire, and is *remote*-sense really necessary?

  • Christmas!

    esot.eric08/30/2019 at 23:58 4 comments

    Quality [and quantity!] to be respected.

    Forgive me if I go "radio-silent" for some time...

  • Great idea "PCB ruler"

    esot.eric08/26/2019 at 22:59 2 comments

    Hard to tell what's all on it from the tiny picture, but handy tool, potential business-card, and just a tiny bit more design effort and a pair of tinsnips, or a utility-knife and table-edge, or snap-perforation, could make it quite useful for prototyping, as well.


  • Home-made diode/transistor

    esot.eric08/26/2019 at 04:31 4 comments

    It really seems to me there must be ways to make transistors with household supplies/tools... yahknow, without a microscope or nasty chemicals or special oven or, most importantly, pre-fab semiconductor-materials like silicon wafers. Here's where my research led...


    metal rectifier is an early type of semiconductor rectifier in which the semiconductor is copper oxide...They were used in power applications to convert alternating current to direct current in devices such as radios and battery chargers.
    Metal rectifiers consist of washer-like discs of different metals, ...copper (with an oxidelayer to provide the rectification) ... interspersed with aluminium discs (which were often of a larger size, to provide cooling).
    The principle of operation of a metal rectifier is related to modern semiconductor rectifiers (Schottky diodes and p–n diodes), but somewhat more complex. copper oxide [is a] semiconductor, in practice doped by impurities during manufacture

    [Hey, i've got some of these!]


    Copper(II) oxide is a p-typesemiconductor, with a narrow band gap of 1.2 eV.
    It can be formed by heating copper in air at around 300 – 800°C:
    2 Cu + O2 → 2 CuO


    "In the history of semiconductor physics, Cu2O is one of the most studied materials, and many experimental semiconductor applications have been demonstrated first in this material:

    [A bit of confusion regarding *which* copper oxide, carries forth later-on as well]



    Shockley's research team initially attempted to build a working FET, by trying to modulate the conductivity of a semiconductor, but they were unsuccessful, mainly due to problems with the surface states, the dangling bond, and the germanium and copper compound materials. In the course of trying to understand the mysterious reasons behind their failure to build a working FET, this led them to instead invent the bipolar point-contact and junction transistors.

    Hah! So mosfets were the goal in the first-place! Then failing that, the bjt was born as essentially a mistake... then twentyish years later the technology existed for the *original* goal.


    The problem with the realization of a working metal-insulator-semiconductor (MIS) field-effect transistor (FET) is the interface between the semiconductor and the insulating dielectric. This was obviously also the case with the experiments on the semiconductor cuprous oxide (Cu2OCu2O), which was historically used as a rectifier. John Bardeen (double Nobel prize winner in Physics) found that this problem is caused by interface states at the insulator/semiconductor interface which hinder the penetration of the electric field into the semiconductor thus preventing the modulation of a mobile charge layer (conducting channel) near the surface necessary for the field effect transistor. In the course of the experimental investigation of this problem with conducting probes on the semiconductor surface (germanium), Bardeen together with Brattain serenpitously discovered the bipolar transistor effect which led to the realization of the point contact bipolar transistor and later the junction bipolar transistor. Very few insulator/semiconductor interfaces have low enough interface state densities to be suited for a field effect transistor. Only in...
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  • Charging the big-guy

    esot.eric08/25/2019 at 04:34 0 comments

    ok, so now there's 40V worth of batteries in series... that's 8 batteries need charging. And, in the last post I explained many hurdles.

    Check this... the DC converters require 36V-min [74 max]... so 40V gives just enough clearance such that my 12V car system can bypass and charge three at a time, leaving 5 totalling 25V + 12V from the car = 37V to keep my loads powered while charging.

    That comes at a slight penalty; now there's not quite as much available maximum power [max current is 2A, but now the voltage has dropped while charging]. Probably not a big deal, really... how often is 37V*2A=74W not going to be enough?

    But, more frustratingly, it means a full charge requires three cycles, 3 batteries each, twice, and two in the last cycle.

    [Hmm, I haven't yet pondered whether they can all be charged simultaneously if the system's not powering a load... 5V 8*2=16A might be a bit to consider].

    OK, anyhow, the idea is simply use a double-pole/double-throw relay for each battery to switch it out of series for the circuit and into parallel for charging, and a diode on the 12V source and another on the last battery in the group handles bypassing.

    Now, if I use that same logic with Even Moar Batteries, we can cut it down to two charging cycles. e.g. 60V cut in half + 12V = 42V > 36V. Now I only need two diodes [less power dissipation], and can charge the whole thing faster, six batts at a time... and I happen to have two 12V->5V 6.8A 'lighter' adapters, which can each charge three of these. And, of course, more runtime, and more max power when not charging. 

    [maybe 10's more reasonable? 25+12=37, although that might require bigger caps, again, hmm... maybe 14! Heck, 74/5=... sure, 15, why not?]

    [BTW, this is getting a bit ridiculous, originally I just planned for rechargeable power for/attached-to a few things like a portable DVD player and TV [4G is expensive, yo!] and the little computer my buddy @Starhawk built me, oh and maybe a reliable power-source for my external backup-hard-drive... and then maybe for my electronics-projects when coding [e.g. #Floppy-bird is still in the noggin'], and then my electronics-projects which needn't coding [e.g. #Incandescent RAM is bubbling up again], and then, maybe the soldering-iron... and... well, it's starting to make a heck of a lot of sense to keep all that power centralized, I see, now, rather'n having a dozen devices each with its own battery[pack], especially since most are not likely to be used simultaneously].

    Now, we still have the problems when I might bump the power-cord outta the lighter-plug, of: batteries which are charging take a quarter-second to turn-on for a load, combined-with: now there's switching time for the relays to put them back in series. So, back to a big ol' cap... two, maybe. But, check this: because our charging-bypass-voltage [12V] is lower than the total voltage of the bypassed batteries, charging the capacitor to the battery-voltage from 12V will *be* the load which will turn on the batteries. So, now, we needn't wait for the capacitor voltage to drop *before* the batteries know to turn on. Thus.... faster start-up. Of course, counteracted a bit by switching relays. But ah well. Besides, now we're talking 12 [maybe one less per charging-bank] relays, anyhow... 's not like we hadn't reached ridiculous-territory long ago.

    [Of course, a reasonable electronics-hacker surely would just install banana plugs in the dashboard and be done with flaky connectors altogether! But there's always portability and dead starter-batts as good reason for this project]

    Relays, yep... break-before-make, yo! [I may reconsider mosfets... but *four* vs one relay?]

  • Charging...

    esot.eric08/24/2019 at 22:43 0 comments

    The not-so-easy part of using all these usb-batteries in series is charging. That's most of half a spiral notebook...

    These aren't simply batteries, they're batteries with a built-in charge-controller on the input and boost converter on the output. [4 terminals!]

    If you thought charging regular batteries in series was difficult or prone to over-charging one while undercharging another, I challenge you to consider what happens with two series charge-controllers when one battery completes its cycle before another. Where's all that series current gonna go? What's gonna keep it dropping 5V?

    Then, to make matters more complex, I really woulda preferred being able run the load off the car-power/batt/alternator *while* charging. No prob, the USB-batts have a 'bypass' mode, putting the input 5V on the output while charging [except, they actually disabled that feature, probably due to current-handling ability either on the charging/output circuit, or in consideration someone might try to charge it and an attached phone at 4A from a 2A port, so you gotta take ground from the input-terminal for both input and output]. And... in series, that means needing separate 5V supplies, put in series, to charge and handle the loads of the now charging batteries in series. Whew!

    But, more difficult is switching from charging to output... [e.g. if i bump the car-plug] the batteries are actually 3.75V, the output is 5V, that means a boost-converter at the output. It takes about a quarter-second to turn on. That means a quarter-second "blackout" when the car power is unplugged. At 5V 2A I calculated needing a 0.1Farad power-cap to prevent blackout. 

    But then it gets even more complicated! The batteries switch to output-mode by detecting a current-draw... if a big ol' cap was put in parallel to the batt when it's off, the *cap* will handle the current-draw until it reaches 3.75V, or slightly lower, then some current will be drawn from the batt, telling it to switch on, then another quarter-second until it's outputting 5V. But, as far as I can tell, it's not 3.75V *direct* from the batt, but 3.75V through a high-value resistor. Then the cap voltage may well drop quite a bit lower than 3.75V by the time the 5V switches on. WHEW!

    I've come up with countless possible solutions for charging, outputting while charging, and outputting without brownout when external power's disconnected, mostly relying on switching them from series-for-output to parallel-for-charging. [That's a lot of switches/relays/fets, and considerations regarding make-before-break, or vice-versa. And the extra time switching.] I've also pondered a few gnarly-hacks, like using two daisy-chained *non*-floating DC-converters [which allegedly don't like sinking current, but *might* actually be capable, since they have both high-side and low-side mosfets, a feedback path, and PWM-control].

    Oh yeah, it gets even better! ... erm, what was it...? 


    Then these floating-output converters came my way... So, dig this... I can "boot" the system from 8x5V batteries, 40V. Then, because the 40V->5V converters' outputs are isolated, and because the batts can charge and output [bypass] at the same time, the 5V from the floating-output converter can be used to charge one of the batts *and* maintain the 40V source it needs in order to create that 5V in the first place!

    It's not free-energy, of course, the power going into charging that batt will be coming from the other batts [at 1/7th the charge, each]. But, now, again, because the system itself is floating [entirely battery-powered] I can *also* bypass two [maybe three] [and possibly also charge] batteries by powering *anywhere* in the battery-chain from the car.

    So, now, 3/7ths of the charge-power is coming from my alternator, 4/7ths from 4 other batteries, which, after, can be fully-charged by depleting the recently-charged batteries by 1/7th charge... or something, here's where it gets a little vague. But, the interesting bit to continue pondering...

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