Electricity 101. I'm taking notes. Teacher(s) Wanted!

[Whistle_Pig]

but now I'm starting to understand the passthrough idea. you're getting the 1000mA from what the computer puts out, making it a 1 amp output, giving it a 1*2.1 = 2.1v output. Not the 5v you would expect with a normal 2amp supply which would give it a 2*2.1= 4.1v output.

<SNIP>

and oh my god, thank you

Yer welcome!

The USB power lines on the computer are a voltage supply. So that means that the current will vary with the resistance, within the limits of the circuit supplying the voltage. As long as you don't exceed the current capacity of the USB port, you'll get 5 volts. I've read various threads here where people are saying that the typical USB port will supply up to 1A, or 500mA. I suppose it varies depending on the design choices made by the people designing the USB controller. That doesn't mean it always delivers 1A, or whatever.

When talking about power supplies, there are voltage regulated supplies, and current regulated supplies, and probably others I don't know about. In the world of vaping, at least in the reading I've done so far, nobody has mentioned using a current regulated supply. A battery is a voltage-regulated supply (or behaves like one -- or we could say a voltage-regulated supply behaves like a battery). So we won't talk about current regulated supplies at all. In the context of a voltage regulated supply, there will be a limit to how much current it can supply while still maintaining a regulated voltage. So be careful when you start talking about using a 1A or 2A supply, and know whether you're referring to a current regulated supply (which we won't talk about anymore), or a voltage regulated supply that has a current limit of 1A or 2A.

 

[Erik_Bear]

The specifications for USB state that the USB system be capable of delivering 500 mA at 5.0 V. Some computers only have one USB hub internally, some have 2, so that would be the maximum draw through the entire system. Keep in mind that keyboards and other stuff use that same power source. That is one reason why all the pass-throughs have a battery in them. The other reason is that the battery is the power regulator for the atomizer.

so it would be .5amps cutting all of those calculations I did in half?

 

[Quick1]

Okay Erik. You've really got me confused now. You're saying that the computer's USB is going to put out 1 amp. I thought the USB port would just give out about 5 volts and the amps you will get would depend on the resistance of your load. So if you have a 2.3 Ohm atty connected on the USB port you'd get 5 / 2.3 = 2.2 Amps.

Can you explain why the USB is always giving out 1 Amp?

A load of a certain resistance will "draw" current at some rate at a given voltage. The power supply has to be capable of delivering the current. If it's not, the resistance is pretty much fixed so the voltage would have to change. I = VR

PC (almost all) USB ports supply 5v and are capable of 0.5 amps.
At 5v a 2.2 ohm load will draw about 2.27 amps.

But the USB port is only capable of supplying 0.5 amps at 5v.
If we use I = 0.5 and R = 2.2 then V would drop to about 1.1
...and that's why USB powered pass throughs don't work so well...

 

[Hoosier]

The USB bus is current limited by circuitry on the motherboard and wall to USB adaptors are usually current limited by its circuitry.

At least you hope your motherboard has current limiting or it can damage the traces.

V=IR can only work if current is unlimited. So if you know your resistance and your maximim current, you'll be able to find your working voltage.

Does that help?

Current determines temp of nichrome wire and wire manufacturers will have charts of size, current & temp of their wire. There is simplification with that as the ramp up time is not factored into these charts.

It seems like you are getting the basics down pretty well.

 

[Erik_Bear]

A load of a certain resistance will "draw" current at some rate at a given voltage. The power supply has to be capable of delivering the current. If it's not, the resistance is pretty much fixed so the voltage would have to change. I = VR

PC (almost all) USB ports supply 5v and are capable of 0.5 amps.
At 5v a 2.2 ohm load will draw about 2.27 amps.

But the USB port is only capable of supplying 0.5 amps at 5v.
If we use I = 0.5 and R = 2.2 then V would drop to about 1.1
...and that's why USB powered pass throughs don't work so well...

There ya go, that hit me on the head I had the formula already, just needed the clarification on the USB ports and whatnot.


And if anyone could, what do watts have to do with vaping? from what it seems like I've drawn, anywhere from 5 to 10 watts is where we are supposed to be at, any higher and our attys die, any lower and they don't do squat. Am I just confusing myself more? Or did I figure something out?

 

[Digs]

So this is why we see people using the 2A or 3A AC USB connectors to plug their USB PT into the wall. That allows them to draw more amps. That makes sense.

 

[Windsage]

Its a voltage regulated supply, so it will try to give you more than the spec. It just might not last long. Also, please keep in mind the protected Li-Ion batteries have current limiters in them, so they limit both voltage and current. If you hook up straight to the USB and an atty shorts out, you could lose a motherboard real fast.

 

[Erik_Bear]

The USB bus is current limited by circuitry on the motherboard and wall to USB adaptors are usually current limited by its circuitry.

At least you hope your motherboard has current limiting or it can damage the traces.

V=IR can only work if current is unlimited. So if you know your resistance and your maximim current, you'll be able to find your working voltage.

Does that help?

Current determines temp of nichrome wire and wire manufacturers will have charts of size, current & temp of their wire. There is simplification with that as the ramp up time is not factored into these charts.

It seems like you are getting the basics down pretty well.

D'oh!
Posted right before I posted mine. So its not the watts, its the current going in that needs to be correct? So amps is what I'm looking for, watts is not?

 

[Scubabatdan]

mA/mAh=C 1.8amps = 1800mA

@2600mAh 1800mA/2600mAh= .7c

thats for the first part of the answer digs had, I'm not sure about the second part.


but now I'm starting to understand the passthrough idea. you're getting the 1000mA from what the computer puts out, making it a 1 amp output, giving it a 1*2.1 = 2.1v output. Not the 5v you would expect with a normal 2amp supply which would give it a 2*2.1= 4.1v output.

But for the atomizers I would be running, 510 3.3ohm are rated at ~3.3ohms, making them run on a normal 1amp supply would have a 3.3*1= 3.3v output, or a 2amp would give a 6v? I think I messed up trying it in my head, lemme grab a pen n paper.


and oh my god, thank you

Only one problem here, the USB port on a computer works in 100ma unit loads, the low power bus can not draw more that 100ma or one load. The high powered bus can only draw a maximum of 5 loads or 500ma at a maximum of 5.25vdc.

So a USB port on a computer cannot supply 1A of power. That is why I use an AC/DC 4.5vdc 1.6A power suppy to a USB adapter for my passthrough.

BTW that is for USB 2.0...
USB 3.0 is suppose to be 150ma unit per load with a max of 6 unit loads on the high power bus = 900ma max for high power bus. Hurry up 3.0 and get here

Dan

 

[Digs]

D'oh!
Posted right before I posted mine. So its not the watts, its the current going in that needs to be correct? So amps is what I'm looking for, watts is not?

The heat of the coil depends on both the current & voltage (P=VI). So you do want to look at Watts too because that’s what will tell you what kind of hit you’re going to get

 

[Erik_Bear]

So a USB port on a computer cannot supply 1A of power. That is why I use an AC/DC 4.5vdc 1.6A power suppy to a USB adapter for my passthrough.

Dan

that was pretty much my plan after reading all of this. Gonna buy a Car to USB and AC to USB with my order from madvapes. 5v @ 2amps seems like its gonna do the trick

 

[Erik_Bear]

The heat of the coil depends on both the current & voltage (P=VI). So you do want to look at Watts too because that’s what will tell you what kind of hit you’re going to get

so theoretically, if somehow I found a supply that ran 10v @ .5amps, I would get the same hit as a 5v @ 2 amps?

So it is the wattage I want to look at? I'm just making sure for my mods that I am planning out so I know where the "sweet spot" is for all of it.


sorry for the double post, was gonna condense it but I can't delete my previous one =\

 

[Digs]

so theoretically, if somehow I found a supply that ran 10v @ .5amps, I would get the same hit as a 5v @ 2 amps?

So it is the wattage I want to look at? I'm just making sure for my mods that I am planning out so I know where the "sweet spot" is for all of it.


sorry for the double post, was gonna condense it but I can't delete my previous one =\

Let's say you're using a 2.3 ohm 510 atty with a 10v power supply capable of outputting .5 amps.

With that resistance, the atty will be trying to draw 4.34 amps (10/2.3), which is way more than the power supply is capable of outputting so it won't work.

The same atty on a 5v power supply at 2amps will try to draw 2.2 amps which is just a little over the maximum so it should work out okay.

 

[Erik_Bear]

Let's say you're using a 2.3 ohm 510 atty with a 10v power supply capable of outputting .5 amps.

With that resistance, the atty will be trying to draw 4.34 amps (10/2.3), which is way more than the power supply is capable of outputting so it won't work.

The same atty on a 5v power supply at 2amps will try to draw 2.2 amps which is just a little over the maximum so it should work out okay.

K, got it. Thanks for the clarification.
If I were trying to make that work I'd need about a 20ohm atty =P which I doubt is out there? XD

 

[Digs]

Also, when I say P=VI. The current I'm referring to is not the maximum current allowed by the source, but the current that is being drawn from the source. I think you're getting the two confused.

 

[Erik_Bear]

Also, when I say P=VI. The current I'm referring to is not the maximum current allowed by the source, but the current that is being drawn from the source. I think you're getting the two confused.

Yep, I was, makes sense now though. Gotta get my head on straight with all of this info flooding in 8-o

 

[Digs]

Hehe.. I hear ya. I'm learning all of this at the same time you are. Eventually we're going to understand it all

 

[Erik_Bear]

Hopefully more sooner than later. I'm planning on modding a bunch this weekend so I want to get most of it down before then.

Now, its time to ask about wiring diagrams. The difference between in-line and parallel and all that other mumbo-jumbo.

 

[Whistle_Pig]

so theoretically, if somehow I found a supply that ran 10v @ .5amps, I would get the same hit as a 5v @ 2 amps?

Sounds to me like you're still linking volts and amps incorrectly.

You can use a 5V supply, or a 10V supply, and the current will depend on the load (for our purposes, we can speak of load purely as resistance in ohms). You can't set up a power supply to always supply 5V @ 2A. (Or, if you can, and I can't imagine how, it's irrelevant to the purpose of mods for vaping.)

Initially, think in terms only of the voltage of the power supply. Divide that by the resistance of your atty, and you'll know how much current the atty will attempt to draw from the supply (be it a battery or a regulated circuit). At that point, ask whether your supply is capable of supply that amount of current.

Don't start out from a position of x volts @ y amps. Start out from knowing the resistance of the atty you're planning on using. Then it's some judgement and reading, because some people report great results with one combination, which other people will dismiss as not having enought throat hit, or whatever. But you'll end up choosing what voltage you want to use for that atty. Once you know that, then you can pick a combination of batteries, or something else, that will supply sufficient amperage.

 

[Erik_Bear]

Sounds to me like you're still linking volts and amps incorrectly.


Don't start out from a position of x volts @ y amps. Start out from knowing the resistance of the atty you're planning on using.

That was my mistake Thanks for the further clarification. That's exactly what i needed to know