Great!!! I am so glad you are going 'back to basics' with this, because these are the parts I don't understand, in red.
So would you say that Volts is actually the equivalent of Volume? That is, the amount of power stored in a given physical space? Or put another way, the VOLUME of POWER contained in that battery? So a 9 volt battery holds nearly 3x the volume of power as a 3.7 volt battery?
The Power is Watts = Amperage times Voltage P = ( I X E )
The energy stored is Watt-Hours.
10 Watt-Hours can be 10 Watts for 1 hour or 1 Watt for 10 hours etc.
If that's the case, what is happening when I set my PV to 4.8 volts on a 3.7 volt battery? This part is where I get lost.... it makes no sense. If an 18650 battery is 3.7 volts, I can't take more than 3.7 volts from it if that's the POTENTIAL power it contains. Can you please explain this??
Magic of electronics.
I wrote this some time ago and it is a little more involved than what you are looking for:
Basic tutorial on current with DC-DC buck and boost converters
I will not go into the internal circuit design itself here as it requires advanced knowledge of power electronics and a little black magic as many have found out during implementaion of a design in the real world. I will however attempt to explain the basics necessary to understand the relationships between input voltage and amperage and output voltage and amperage which is hidden behind a lot of equations that look like alien writings. Bear in mind this is simplified and there are many things I am leaving out about the actual internal workings. Also I will assume the external circuit is properly designed for the application.
To understand these relationships we need to briefly look at how the converters work.
A DC-DC buck converter, also called a down converter, will always have the output voltage lower than the input voltage.
(A linear regulator is a form of down converter but is less efficient as it has to waste all the power in conversion since it does not store energy)
A DC-DC boost converter, also called an up converter, is very similar to a buck except the output voltage will always be higher than the input.
There are various other converters (buck-boost, flyback, SEPIC, Cuk, and variations) that will provide both lower and higher output voltages than the input voltage but are normally found in complete end user devices.
The following will apply to all modern DC-DC converters.
A DC-DC converter briefly stores energy from the input in either a magnetic field or a capacitor. In most all converters it is in a magnetic field so we will look at them. A magnetic field is built up in what is called an inductor that is a coil or that block looking thing with a coil inside it that is normally the largest component in a converter. Once that energy is stored in the magnetic field there is no longer voltage or current, just the magnetic field strength. It then releases that energy to the output. During this release the design of the converter and possibly external components determines what average voltage is present at the output.
What about current? Since voltage and current "disappear" inside the inductor when they create the magnetic field, input current has no direct bearing on output current.
To determine the output current is simple. Divide the output voltage by the load resistance. I.E. if the output voltage is 4 volts and the output resistance is 2 ohms then the output current is 2 amps. (4V/2O=2A)
To determine the input current however you must know the input or output wattage, converter efficiency at the working parameters, and the input voltage. Since we will normally know the output wattage we take output voltage times output current. Using the above example 4V X 2A = 8W of output wattage. Converters do not have the same efficiency through their range so normally you will have to look at charts provided by the manufacturer in the data sheet to determine the efficiency at the input and output we want to figure it for. For simplicity we will say that in this case the converter is working at 90% efficiency. So if we are getting 8W out of it there must be 8.9W going into it. 8W / 90% = 8.9W (rounded up to the 1/10th watt). To figure the current input we take the input wattage divided by the input voltage. Lets say we have 8.2V input (2 cells at 4.1V each). 8.9W / 8.2V = 1.1A (again rounded up)
So you can see that while we are getting 2A out of the converter we are only drawing 1.1A from the batteries. However the input current will rise as the batteries discharge. Lets say our batteries have fallen from 8.2V total to 6.4V total. Thanks to the converter we still have 8 watts of output which means we still have 8.9 watts of input (assuming efficiency has not changed). But for the converter to get that 8.9 watts from the batteries it has to draw more current since we are giving it a lower voltage. 8.9W / 6.4V =1.4A. In both cases our batteries are seeing a lower amperage than the output is providing.
Now lets look at a boost converter. The math is the same. Lets say we still have 8W output at 4V and our input voltage is now 3.2V (1 cell) and our efficiency is the same at 90%. So the input wattage will still be 8.9W but the input current will not be the same. 8.9W / 3.2V = 2.8A. Now our battery is seeing more amperage than the output is providing.
As a general rule of thumb a buck converter will always draw less amperage from the batteries than the output is providing and a boost converter will always draw more amperage than the output is providing. A converter that will do both more and less voltage out than in will as a rule of thumb do the same as a buck if it is giving less voltage out than in and the same as a boost if it is giving more voltage out than in.
You should always look at the lowest battery voltage the converter will see when determining what the maximum amperage the draw will be on the batteries.
Most all modern DC-DC converters will have a thermal shut down mode. Internal components will determine if the converter is getting too hot and will shut the converter down. Converters need to shed heat. This heat build up is determined by load, duty cycle (how long they are on compared to off), and how much they can shed this heat. If driven hard and a long it is important to not keep air from being able to convey this heat away or thermal shut down may occur.
Most all modern DC-DC converters also incorporate what is normally called the "hiccup" mode. This is incorporated into most all modern DC-DC converters. When a set point is reached (determined by the manufacturer) where the electronics inside the converter are not able to provide the current out that you are trying to draw from it, the converter will go into hiccup mode. The converter will output the most it can for a very brief time then output nothing for a longer period of time (a short duty cycle) then start over again. You can see where the name hiccup came from. This is to prevent damage to the internal components. The point this happens is normally refereed to as the "inception point". The inception point will always be higher than manufacturers rated amperage for the converter. If not the converter could never reach the rated amperage without cycling into and out of hiccup mode. Normally a manufacturer will set that point at 110% to 140% of rated amperage. This is determined by the manufacturer and will be below what destructive testing has shown will damage the internal components. The inception point will normally be included in the manufactures data sheet.
Most manufacturers will also include a warning that a fuse should be used to protect the converter and power supply. This is because hiccup mode will only protect the conveter for a short period of time against a dead short. In some cases 1 second or less. If the converter fails in a short mode instead of open mode the batteries will be directly connected to the load since the majority of converters are "non isolated" and the ground for the input is also the ground for the output.
Ok, this is pretty straightforward, and this is what is measured by Watts right? How much is flowing at a time, VOLUME of electricity flowing, or the SPEED of the electrical flow? These are 2 different things... which is it?
Speed is measured by the electromagnetic wave and not the movement of electrons and is dependent on the speed of light times the dielectric constant of the medium the wave is moving through and is not at all pertinent to what we are discussing and would take a long time to explain.
How much is flowing is Amperage.
How much is flowing (Amperage) times the pressure (Voltage) is Wattage.
Ok, this makes sense... it's the amount of RESTRICTION you are placing on the flow. Ie, a 1mm opening, 5 mm opening, etc, affects the SPEED you are releasing the electricity from the supply, or the voltage pool.
Not speed but how much is flowing (Amperage)
I have never seen the term 'conductance' used in vaping. Where is it used, and what is it relevant to, when we use primarily resistance for gauging coil size/length? Is this applied to the battery itself? Perhaps the meaning of the C designation (below)?
I've never seen it used in vaping either, but was just mentioning it is the reciprocal of resistance and is equivalent to opening up the faucet.
Ok, I got this, with above caveats and request for further information.
Ok, this is also straightforward.
Ok... I will study this, as it's the meat of the matter. I really appreciate your explaining it yet again (because this probably comes up constantly in newbie discussions). Could you help answer some of my questions above tho? Thank you very much!!
.
LOL! Sok, I'm interested, and am going to try to absorb this... you have NOOOO idea how hard that is for me, because like physics, electricity defies the laws of logic. um. (I didn't like calculus either... for me, 2+2 should always equal 4, not 7).
But I'll get there
glassgal- there will be a test on all of the above which you need to pass before getting your dibi...
.
I bet your like my dad aren't you?
He gave a 50 question true / false pop quiz to his advanced accounting class with all true questions.
No one got 100%
?
That's because you don't visually look at everything as a fibonnaci sequence... which is logical
I'm trying to review what I know about mech mods, and try to fill in the blanks... would appreciate any help
Ohms Law - where it would apply to mech mods, it's that your battery basically drops off of the amount of voltage delivered over a period of time, so that your vape is going to diminish in quality and quantity over the life of the battery in it. What else about ohms law applies?
battery safety- don't overcharge, tear it out of the unit as soon as it starts feeling hot and get it away from yourself and your unit (but don't throw it out a closed window because that can get expensive).
proper cleaning and care of contacts - swish all contacts down with 99% alcohol once a week or so, and maybe once a month, use a nice spray electronic cleaner on it if you have any juice leaks. If not, squirt with canned air.
That's the only thing to know about mech mods? Nothing else?
How about the advantages of sub-ohm? If sub-ohm vaping requires a power draw as high as 30 watts... isn't that a super short battery life, requiring more frequent battery purchases and more frequent battery changes/charging? What are the vaping advantages for doing it then, besides the need to spend more money, buy/make more juice, and do more busy work (recoiling, recharging batteries, refilling with juice).
Given the above, would it not be more practical to build over .8 ohm coils at all times? What is the taste difference for sub-ohm coils that people would sacrifice battery life for? Thank you!!
.
That's because you don't visually look at everything as a fibonnaci sequence... which is logical
