Is This Possible? A 1.5 Volt Battery Lighting Up A 3 Volt LED - Post #3
My eldest son on his recent photos, how time flies. May you fly high josh, may you not stop seeking what is good and Godly. I have this feeling one day you will tell me and mom “IT’S TIME FOR ME TO SAVE THE WORLD DAD……...”
WHAT REALLY HAPPENS WHEN A JOULETHIEF IS RUNNING
As this is my third post, I am really hoping that you have read my first two posts here so as to have a better picture of what I have to say here:
https://steemit.com/introduceyourself/@lightingmacsteem/lighting-for-everyone-the-steemit-way
https://steemit.com/offgrid/@lightingmacsteem/7mcde8-lighting-for-everyone-the-steemit-way
Try putting an LED directly to a 1.5 volts battery, positive to positive and negative to negative of course. It may take us 10 lifetimes, and I assure you the LED will not light up this way. But place the LED the way as presented in the joulethief circuit, and it will light up!!!
So how does a joulethief, at only 1.5 volt battery source, able to light an LED that needs 3 volts to light up? What really happens inside and around the circuit? I need of you as the reader to remember the things as I describe here in this third post as this will be the very basic principles that we will be using as we move on to higher voltages and possibly more powerful use of this wonderful circuit. Consider the drawing below as per the last joulethief drawing, but this time I added a flow of current, the blue line indicates the current flow when the transistor is at “close” position while the red line indicates the current flow when the transistor is at “open” position, here it is again:
During the blue line “activation”, we can safely conclude that the battery is “draining” but at the same time the coil (inductor) is “charging”. I say activation during this stage of the run because it is when the transistor is at closed position that this circuit activates initially. Wait a minute, inductor charging? Well, I just use that term to simply tell everyone that an inductor is “saving” power at this point, and when I say power, I meant voltage and current at levels that can be used if only we know how to manipulate it properly, or should I add the word honestly, as oftentimes this area of an inductor’s function is somewhat suppressed by the mainstream scholarly figures. Let us always remember that an inductor or coil can and will “save power” as long as current is passed thru it, this is always its behaviour and always will be. There will be no suppressing of this now, at least in this post. I have this feeling that I will discuss a bit about the so-called flywheel diode configuration to a motor, a circuit that Engineering books tells us is best. Actually, it is only best at wasting “real” power from a motor, but for now just consider this as a trivia. Let’s move on.
So what happens after the “blue” line activation? I would like to use the term “healing” during the time when the “red” line de-activation happens. Why healing? Well, electricity as it turns out always heals itself, meaning it wants to go back to the state where it was before it was activated. Imagine a rubber band, stretch it as far as you can without destroying it, then release one end suddenly. If you will focus a videocam on this exact scenario, you can clearly see that the rubber band would like to return to its original length, meaning, it wants to “heal” itself. But surprise of all surprises, and if your videocam is fast enough, you can also see that when the rubber band is trying to return to its old state of not stretched, you can see that it “overshoots” first its own original length before it slowly returns to its final and true length. This is exactly the same with an inductor “healing” itself on electrical terms. When the “red” de-activation current happens, the inductor is releasing its stored power, and like the rubber band it will do this “overshooting” its initial state first before finally resting. So how much of an overshoot does this joulethief inductor really accomplished? Well, as the source battery is at 1.5 volts, clearly the overshoot voltage is more that 3 volts because it will light up the LED which is supposedly to light up only on 3 volts. Remember my wind example of things unseen and yet still understandable? How on earth can there be any other reason for the LED to light up if the inductor is not lighting it at 3 volts? So directly speaking, the inductor is the one that lights up the LED, not the battery as it is only at 1.5 volts.
LESSONS LEARNED SO FAR
What do we have here as of this point, let me enumerate the very clear and good-to-experiment-more facts:
-The inductor will always heal itself, but everytime it does this, it always overshoots the battery’s voltage, the battery being the source power in every “activation” cycle
-The battery is not involved directly in the lighting of the LED, it is the inductor
-The overshoot of voltage is “captured” by the LED, we are sure of this because it lights up; this is probably one of the most important facts here, that whatever and wherever the said current comes, whether from the inductor itself or from another dimension is not important, what is important to us is that the LED is able to capture such
-More voltage means more power, if only we know how to manipulate it; remember the inductor overshooting always here, at the very least a thousand times per second
-The cycles per second (meaning how many times per second) when an inductor overshoots itself is more than a thousand times, probably more, and every cycle means an overshoot of voltage; just let that sink in for awhile, because we may have a situation here where you drop 1 dollar to an ATM machine, and each time you do it will let you withdraw 2 dollars!!! Well, just a possibility for now as far as the stated facts are concerned
-The basic joulethief is just the simplest arrangement for this; I can not count the variations for this wondeful circuit, all having huge potential, that is why finding the best approach is my main aim here to share with everybody
-In the forums, even a “dead” 1.5 volt battery lights up an LED for days; days, from a dead battery
LET US IMPROVE THE JOULETHIEF FURTHER
If these are the clear facts, then we will have lots of things to make to “improve” on such, let me enumerate just some:
-I can replace the LED with a component that just stores the charge as given off by the inductor, this way I can use it for other much meaningful purposes
-I can increase the voltage of the source battery (say from 1.5 volts to 12 volts), and as observed, even at twice the source voltage overshoot, I am assured that I will have 24 volts as given off by the inductor
-What if I will re-arrange things from this simple circuit? See what happens
-What if I will add components to this simple circuit? Perhaps an antenna? Perhaps a connection to the ground? See what happens after
-What if I will “store” the voltage on one circuit, then use the stored voltage to power another circuit, will it increase the overall power? See what happens
-What if I will return the stored voltage to the battery source? Will it replace the “drained” power?
-What about if I will combine all ideas into one circuit?
Stay tuned, I can guarantee an interesting ride. If some electronics enthusiast is reading this now it may be time to be ready to replicate, I urge you to do this good help, for this world’s sake.
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STOPThat is actually a very interesting read! I learned something new. Thank you! Do you think Tesla is using this thinking in designing their batteries?
Also, for how long does the overshoot actually light the LED? A few seconds or less I assume?
Thanks for the upvote!!! Actually Tesla is way way far ahead, we are just getting a glimpse of it now. Yes, he was thinking the same. The overshoot is but a blink of a second, so so so fast. The LED is actually blinking on and off, but a thousand or even more per second, your eyes can not really grasp it anymore, so your brain is the one telling you THAT IS A STEADY LIGHT RIGHT THERE. Stay tuned for more posts, we can fill in this whole year just studying this and replicating. Regards.
That is so neat! Efficient lighting :)
Thanks man, you encourage me with your good words.
I need to reply to this again as I first thought you meant TESLA the scientist, I believe you meant TESLA the elctric car company here. I believe they are deploying lithium ion batteries, but I really suspect it is more of an "unknown" advanced technology for SUPERCAPACITORS that they are using. I may be wrong though.
We will discuss SUPERCAPACITORS in the later part of my posts surely. Regards.
Isn't the oscillation of the coil drawing the 'healing' power from the battery? IIRC, volts, watts, and amps are in a relationship such that volts x amps = watts. So, when the transformer (coil) is opened such that the electrons are released, the voltage sufficient to power the LED is available.
When the transformer is reconnected to the battery by the transistor, then it draws power from the battery, and the LED does not have sufficient voltage.
I reckon that the use of two batteries in this circuit should last twice as long as the single battery, and this would indicate that no ergs of energy are entering the system from somewhere besides the battery, but that the transformer is, well, transforming the electricity into 3 volts for half the time the battery is connected.
Have you done this experiment?
Wow!!! One of the most technical question I have received so far, I can see you are a learned man Sir, I salute you, and hopefully you have read already my previous two posts. Actually, when we try to understand the healing part of the cycle, we need to understand as to why it heals in the first place, because healing is the circuit's reaction when it is energized initially. There is always a reaction for every action, equal and opposite right?
I see this circuit's healing cycle as its reaction, but reaction to what? The inductor is just reacting to the fact that it is always being "nudged" by the battery's voltage/current/wattage components (meaning electric), and every time the battery nudged the inductor, it will always go back to its original state and position after the nudge, hence the magnetizing (the battery nudged the inductor firstly) and de-magnetizing (the inductor going back to its original state before it was nudged) is but the inductor's reaction to the battery's action.
You are very much correct that more batteries means more time to operate. I have done this experiment for the last six years and counting on the same and simplistic joulethief circuit, I just modified it so many times and it is when things will get exciting when we do that. If you can replicate this Sir when we will reach that point, I will be privileged as we can maybe teach the world how to do this as well. Regards.
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STOPfollow and vote
https://steemit.com/@rohit26
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STOPsteem-untalented
This is a sample reply for now.
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Thanks man! I am really new here, and I plan to post daily. So how did this resteem benefit my articles? Thanks for the goodwill.
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This is just a sample reply from the author @hippiemom