Nice one, that's a proper case right there.... One thing I noticed is the fire button is on the opposite side of the atty. All other box mods have the fire button and atty on the same side of the box.
Sigelei 100 watt dual 18650 box mod
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Aaaaand I find out that the one I was supposed to be getting from my prebuy isn't going to happen because sigelei thought it was a good idea to send 10 black units instead of the requested 5 black and 5 silver. This better not turn into the debacle that happened with Cana colored units.
I don't think it's going to be that bad. Yesterday vapercompany updated their website listing of the sigelei 100w with new pictures after they received their first ten units. And it appears they received both colors because the new pics of the black and silver mods look like they were taken by them.
At the same wattage the 100w with 2 batteries will last twice as long as the 50w with one.
Come on people, conservation of energy.
Only if the battery's are in parallel would you get double the battery life and amps. These are in series, so all it doubles is the voltage.
No. Whether it's series or parallel you have twice the battery. Either you have ~5000mAh @ 3.7v or 2500mAh @ 7.4v. Either way will do the same amount of work. If it wired in parallel it would roughly be drawing 30 amps @ 100w and that would split between two batteries (15A each). If the cells are in series, the higher voltage would only require 15A from the pair in series. Same work done, just a different arrangement depending on what voltage the board requires to operate.
OK thanks, that make's sense to me, I was all excited about having twice the battery life with my curent setup, then I saw pictures of it in series and was worried because I thought it had to be in parallel to achieve that.
Are you certain the amp draw is the same with parallel and series set ups in a regulated device? I always thought it would be but on another thread recently I went back and forth with someone who convinced me the amp draw is half in parallel than in series...
The following is only some of the responses mostly to my questioning about amp draw being half in parallel than it is in series with a regulated mod....
The run time is ONLY the same if it's a mech., keeping ONLY wattage constant.
In a mech, you CANNOT have both constant wattage and resistance for both series and parallel. It's just physically and mathematically impossible.
It is ONLY in a regulated mod that you can keep output voltage and resistance constant (and this mathematically and physically keep wattage constant).
However in that case, current drawn from parallel is half that of series, ie lower strain and longer run time.
I'm not sure I understand the second question. You CANNOT control the voltage output of a battery. Also again, for a mech, you cannot fix both wattage and resistance.
Let me answer the question in 2 parts:
1) the chip doesn't not control how much voltage is given by the batteries. Regardless of whether the input voltage is 8.4V or 4.2V, the chip takes it in, and uses it's circuitry to output the required voltage.
For example, if the required output voltage is 5V, in a series set up, the chip will take all 8.4V, and use it's buck circuitry to "chop" the voltage, dumping out only 5V. (This is obviously an over simplification).
2) the significance of the cells not seeing the load is this: the inherent total voltage does not affect the current provided per cell. Think of it this way.
The batteries tell the chip: here, I have 8.4V.
The chip says "dude. But I only need 5V. Ah whatever I'll take it anyway."
The chip looks at the load and says "ah. The resistance is 1 ohm. So I need to give 5amps to the load. Brb"
The chip goes back to the battery: "alright I need you to gimme 5amps. I'll use 5V out of the 8.4V you gave me, to push this 5amps of current to the load plxkthnxbye"
The batteries give the 5amp.
In parallel, the batteries say "oh well. We will each give 2.5amps each for a total of 5amps"
In series, the batteries says to each other "dude. If I'm giving 5amps, you're giving 5amps too. "
Based on the series rule, the 2x 5amps from both battery are not additive. It's just how it is. So the net is still a 5amps
A good analogy: think of the universal voltage converters. If you use it at a country with 230V, the wall outlet is STILL outputting 230V. You can't change that.
What does happen is that the voltage converter takes all 230V, and then chops it up to output 110V.
Then depending on how much current your appliance needs, it'll pull a specific amount of current from the voltage converter, which in turn pulls that specific amount of current from the wall socket.
In short,
Voltage from primary source = constant and cannot be changed
The load only sees a voltage from the intermediate (the chip)
The voltage at the intermediate is different from the voltage at the source.
The load draws a specific current from the intermediate, based solely on the voltage output of the intermediate, which draws it from the primary source.
As such, the voltage of the source is irrelevant. It is simply a current source.
Consequently, strain on parallel is less than series.
Lol ok last one. I see where the confusion is now. If you read my previous post and don't get it, here's smth else.
The confusion lies in this: you're thinking that if the chip needs to output 5v (as specified by the user input), then what the chip does is it will ask for 5V from the batteries.
But that is not the case. There is no way to control the voltage output from a source.
What the chip does is this: it knows that it has to output 5V to a load. So it takes ALL 8.4V from the batteries, and using it's own circuitry, chops it down to 5V before outputting it to the load.
This is why the voltage of the source is irrelevant. If a user requests X voltage, the chip still takes ALL voltage from the source. The only difference is that it takes all that voltage and increase/decrease it before throwing it to the load. Then after that based on ohms law, it draws the required current to power the load based on the output voltage and resistance.
If the user specifies a voltage output, then in a series circuit, the 2 4.2V gets lumped together to give a total of 8.4V (this part you got.). When the chip "discards" the excess 3.4V and uses the remaining 5V, it doesn't know who contributed to the 5V. As such it's inaccurate to say that the regulator is using 2.5V from each battery, because all it did was to take 5V from one large pot of voltage.
The next thing is that if the load pulls 10A, then BOTH batteries will have to output 10A because they are in series. It is not 5A per battery, because there isn't sucha a thing as "the regulator is using 2.5V from each of the 2 batteries". The voltage of the batteries are fixed, and are irrelevant to the current being pulled.
In series, given a specific amp draw (eg 10A), be it by a load directly or with a chip intermediate, BOTH batteries must supply 10A, because current in series is not additive.
The only difference between a mech and a regulated mod is
In a mech, the total voltage (8.4V) directly affects how much current is drawn based on ohms law. (Battery voltage and load)
In a regulated mod, the current draw is based directly on the output voltage from the chip and the load.
The run time is ONLY the same if it's a mech., keeping ONLY wattage constant.
In a mech, you CANNOT have both constant wattage and resistance for both series and parallel. It's just physically and mathematically impossible.
It is ONLY in a regulated mod that you can keep output voltage and resistance constant (and this mathematically and physically keep wattage constant).
However in that case, current drawn from parallel is half that of series, ie lower strain and longer run time.
I'm not sure I understand the second question. You CANNOT control the voltage output of a battery. Also again, for a mech, you cannot fix both wattage and resistance.
Let me answer the question in 2 parts:
1) the chip doesn't not control how much voltage is given by the batteries. Regardless of whether the input voltage is 8.4V or 4.2V, the chip takes it in, and uses it's circuitry to output the required voltage.
For example, if the required output voltage is 5V, in a series set up, the chip will take all 8.4V, and use it's buck circuitry to "chop" the voltage, dumping out only 5V. (This is obviously an over simplification).
2) the significance of the cells not seeing the load is this: the inherent total voltage does not affect the current provided per cell. Think of it this way.
The batteries tell the chip: here, I have 8.4V.
The chip says "dude. But I only need 5V. Ah whatever I'll take it anyway."
The chip looks at the load and says "ah. The resistance is 1 ohm. So I need to give 5amps to the load. Brb"
The chip goes back to the battery: "alright I need you to gimme 5amps. I'll use 5V out of the 8.4V you gave me, to push this 5amps of current to the load plxkthnxbye"
The batteries give the 5amp.
In parallel, the batteries say "oh well. We will each give 2.5amps each for a total of 5amps"
In series, the batteries says to each other "dude. If I'm giving 5amps, you're giving 5amps too. "
Based on the series rule, the 2x 5amps from both battery are not additive. It's just how it is. So the net is still a 5amps
A good analogy: think of the universal voltage converters. If you use it at a country with 230V, the wall outlet is STILL outputting 230V. You can't change that.
What does happen is that the voltage converter takes all 230V, and then chops it up to output 110V.
Then depending on how much current your appliance needs, it'll pull a specific amount of current from the voltage converter, which in turn pulls that specific amount of current from the wall socket.
In short,
Voltage from primary source = constant and cannot be changed
The load only sees a voltage from the intermediate (the chip)
The voltage at the intermediate is different from the voltage at the source.
The load draws a specific current from the intermediate, based solely on the voltage output of the intermediate, which draws it from the primary source.
As such, the voltage of the source is irrelevant. It is simply a current source.
Consequently, strain on parallel is less than series.
Lol ok last one. I see where the confusion is now. If you read my previous post and don't get it, here's smth else.
The confusion lies in this: you're thinking that if the chip needs to output 5v (as specified by the user input), then what the chip does is it will ask for 5V from the batteries.
But that is not the case. There is no way to control the voltage output from a source.
What the chip does is this: it knows that it has to output 5V to a load. So it takes ALL 8.4V from the batteries, and using it's own circuitry, chops it down to 5V before outputting it to the load.
This is why the voltage of the source is irrelevant. If a user requests X voltage, the chip still takes ALL voltage from the source. The only difference is that it takes all that voltage and increase/decrease it before throwing it to the load. Then after that based on ohms law, it draws the required current to power the load based on the output voltage and resistance.
If the user specifies a voltage output, then in a series circuit, the 2 4.2V gets lumped together to give a total of 8.4V (this part you got.). When the chip "discards" the excess 3.4V and uses the remaining 5V, it doesn't know who contributed to the 5V. As such it's inaccurate to say that the regulator is using 2.5V from each battery, because all it did was to take 5V from one large pot of voltage.
The next thing is that if the load pulls 10A, then BOTH batteries will have to output 10A because they are in series. It is not 5A per battery, because there isn't sucha a thing as "the regulator is using 2.5V from each of the 2 batteries". The voltage of the batteries are fixed, and are irrelevant to the current being pulled.
In series, given a specific amp draw (eg 10A), be it by a load directly or with a chip intermediate, BOTH batteries must supply 10A, because current in series is not additive.
The only difference between a mech and a regulated mod is
In a mech, the total voltage (8.4V) directly affects how much current is drawn based on ohms law. (Battery voltage and load)
In a regulated mod, the current draw is based directly on the output voltage from the chip and the load.
That's what I thought, but now I'm not sure. The board may be doing something I'm not taking into consideration.
Positive.
Forget the voltage on the atomizer side of things. The chip wants to deliver 100 watts. For simplicity disregard the efficiency of the board.
Power (watts) = Volts x Amps
If you average 3.7 volts the chip is asking for 27.03 amps (split between two cells) to deliver that 100 watts.
In series (7.4 volt average), only 13.515 amps are required to deliver 100 watts.
In series you have double the voltage but half the capacity, so it equals out.
With mechs, when you double the voltage you double the power, so battery life stays relatively the same.
These last few posts have been super informative, thanks everyone 
I'm still kinda confused about a few things, and this Sigelei will be my first high wattage VV/VW device. I'm used to running coils around .5 ohms on a mech, and I've always calculated current draw (for the purpose of ensuring battery safety) using ohms law with the coil resistance and voltage of 4.2, assuming a fully charged battery. So I never really looked at wattage, as it was just a function of resistance and voltage and didn't really mean much in a mech. But I'm confused a bit for what it means with a VV/VW device. Does a VV/VW device only either function as VV or VW - so if I'm adjusting the voltage, the wattage will change accordingly, and vice versa? That seems like it must be the answer, because ohms law is, y'know a law... but this is all still pretty new to me.
I'm still kinda confused about a few things, and this Sigelei will be my first high wattage VV/VW device. I'm used to running coils around .5 ohms on a mech, and I've always calculated current draw (for the purpose of ensuring battery safety) using ohms law with the coil resistance and voltage of 4.2, assuming a fully charged battery. So I never really looked at wattage, as it was just a function of resistance and voltage and didn't really mean much in a mech. But I'm confused a bit for what it means with a VV/VW device. Does a VV/VW device only either function as VV or VW - so if I'm adjusting the voltage, the wattage will change accordingly, and vice versa? That seems like it must be the answer, because ohms law is, y'know a law... but this is all still pretty new to me.
These last few posts have been super informative, thanks everyone
I'm still kinda confused about a few things, and this Sigelei will be my first high wattage VV/VW device. I'm used to running coils around .5 ohms on a mech, and I've always calculated current draw (for the purpose of ensuring battery safety) using ohms law with the coil resistance and voltage of 4.2, assuming a fully charged battery. So I never really looked at wattage, as it was just a function of resistance and voltage and didn't really mean much in a mech. But I'm confused a bit for what it means with a VV/VW device. Does a VV/VW device only either function as VV or VW - so if I'm adjusting the voltage, the wattage will change accordingly, and vice versa? That seems like it must be the answer, because ohms law is, y'know a law... but this is all still pretty new to me.
Correct. Its either/or. If you change the wattage to 100w, the device will figure out how many volts it needs to deliver 100w. If you adjust voltage, you decide how many volts you want and the wattage is a by product of whatever you chose. Its impossible to go adjust the voltage and then switch over and also adjust the wattage. Whenever a change is made to either, the other is affected and changes accordingly.
I do not have one, but I imagine that the Sigelei 100w is only variable wattage. Most high powered devices are nowadays.
Thank you very much mitro! I copied and pasted your post but he doesn't agree. His post is down below. Will you please take a look and tell me what you think?
Hi!
I'm pretty certain that his argument is incorrect.
The premise if his argument is this:
If the user sets a required voltage of 5V, the chip will ask for 5V from the batteries.
That is simply not true. You CANNOT dictate the voltage of an independent voltage source.
A very simple demonstration and proof that it is incorrect:
When you go to asia, you cannot use a standard 110V appliance. Doing so will fry your appliance.
What you do is you use a voltage converter, such that the voltage TO THE APPLIANCE is 110W. How does that happen?
Simple. The cheapest form of voltage down converter is what is known as a "chopper" in layman's term. It takes 230V from Asia's wall socket, chop it down to 110V, and then outputs it to the appliance.
So, FACT:
1) the appliance does indeed receive 110V
2) the wall socket in Asian is STILL outputting 230V
3) the first 2 facts then leads to the simple conclusion that with an intermediate (whether it's a chip, converter etc), the source voltage is irrelevant to the voltage that the load sees.
Blazinga.
I just wanna clarify that I do t think that fella is lying or anything. In fact I see that he has a good grasp of basic ohms law. BUT I'm not sure he understands exactly how a chip functions in regulating input and output voltage and current.
ETA: another glaring mistake in his argument.
He states that forget the voltage at the side of the load, and instead look at the voltage of the batteries on the side of the batteries.
That is against ohms law. For a voltage to drive a current across a load, the voltage source MUST share 2 voltage nodes with the load. The chip (being in the middle) prevents that. Instead, ONLY the chip shares 2 voltage nods with the load. As such, contrary to what he is saying, it is exactly the output voltage of the chip that matters, NOT the batteries.
To confirm, simply google current in a series circuit.
You will immediately see what I mean by sharing 2 voltage nodes: the positive end of the battery will lead directly to one end if the resistor, and the negative end of the battery will lead directly to the other end of the resistor. Because and only because they share 2 common voltage nodes, that's why it's voltage directly drives current across the load
So in simple terms, will the 2 battery mods have better battery life or are those of us that have ordered one specifically for that, wasted our money? I would be mainly using a 1.5 ohm coil kayfun
You guys are too smart for your dam good lol. So who is right? You both seem intelligent in the matter. Guess I have to do my own research. Lol
I ordered from vnw I believe im in the 120-140 range so I guess it will be a few more weeks. Has anyone recieve thiers from vnw yet?
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