Ego boooster!!!

[vocr]

vocr - Re-read your review. All positive for eGo Booster (Minor exception relating to LR attys). GrimmGreen review here quite positive for eGo Booster. Your time use numbers suggest that the calculations shown here http://www.e-cigarette-forum.com/forum/happyvaper/205247-ego-booster-battery-life.html are conservative. Seems like for those that can afford it (and can find it in stock), it is worth a try. I think I will try it and see for myself if it seems worth the price. But still don't quite understand why it will not work with Riva type batts or, in your case, Dealextreme ego batts.

That I cannot answer JW50. Might be because the 4.2v batts don't have PWN? I really don't know. More fuel to that question, I just got 1000mah Echo 4.2v batts and they do not work either. GIMike's analogy seems to make sense, but since I do not fully understand, I have no clue.

Truth is I don't understand that math involved in that battery calculation. However, the results from the math would be "perfect". In my anecdotal testing, it is very not-perfect . My vaping usage, I am sure, varied per test cycle. Possibly even by large margins. I logged what I could. My intention was to try and remember to note in my phone when I put down the PV for a long period of time or did an extra long chain vape session, but it turned out to be unrealistic for me to do that

 

[MasterofChaos]

JW50,

Remember, the capacity of a battery (whether primary or rechargable) is measured in charge, not watt-hours or whatever. The unit of measure of charge is Coulombs. mAHr is also a measure of charge (time multiplied by current), although offhand I don't know the relationship between mAHr and Coulombs. The time between charges will be determined by how many of those mAHr you use up over a given period of time.

The watts out of the battery really have nothing to do with it, and it would actually be very difficult to calculate the number of Watt-hours you get out of a battery since the output voltage would always be changing. It would be a dynamic relationship dependent on the profile of the voltage drop.

Here's how I get the 50% increase in power: An eGo battery, even though its duty cycling, puts out about 3.7V during the on cycle. If you increase that to 4.7V, and the duty cycle doesn't really change (which you can see on the 'scope in the tech video), then the power relationship is:

Old power = (3.7)^2/R (R is the atty resistance)
New Power = (4.7)^2/R

The ratio of new power to old is:

=((4.7)^2/R) / ((3.7)^2/R) - The R's cancel out
= 22.09/13.69 = 1.613 - Remember, this is a ratio, so it does not have any units.

Thus the power is increased by about 61%, which is actually a little higher than the 50% I quoted. This is where I got the 50% number from (I didn't put it in the battery life calculations because I thought 50% was more straighforward). This actually increases the current draw by about 27% rather than 25%, so I guess I was a little bit low in my estimate.

 

[MasterofChaos]

JW50,

If you look at my Post #24 on this thread, it explains in detail why the booster won't work on a Riva, or any of the unregulated or 3.7 volt fatt-batt devices. It comes down to the fact that the Booster tries to keep the output voltage constant, rather than letting it drop when the battery can't handle the load.

And Vocr, I doubt that the booster would work on any 4.2 volt fat-batt devices unless they had an IMR (high-drain) battery in them. The battery would just not be able to supply the current without a big voltage drop and then you would get that run-away voltage drop thing that I described in that post.

That is pretty standard for any Buck/Boost circuit, BTW. If you look at the Provari (I guess the Darwin doesn't count, because it has its own batteries built in) or any other PVs that use a Buck/boost, they all say you have to use high-drain batteries. Its really not at all unique to the eGo Booster.

 

[JW50]

JW50,

Remember, the capacity of a battery (whether primary or rechargable) is measured in charge, not watt-hours or whatever. The unit of measure of charge is Coulombs. mAHr is also a measure of charge (time multiplied by current), although offhand I don't know the relationship between mAHr and Coulombs. The time between charges will be determined by how many of those mAHr you use up over a given period of time.

The watts out of the battery really have nothing to do with it, and it would actually be very difficult to calculate the number of Watt-hours you get out of a battery since the output voltage would always be changing. It would be a dynamic relationship dependent on the profile of the voltage drop.

Here's how I get the 50% increase in power: An eGo battery, even though its duty cycling, puts out about 3.7V during the on cycle. If you increase that to 4.7V, and the duty cycle doesn't really change (which you can see on the 'scope in the tech video), then the power relationship is:

Old power = (3.7)^2/R (R is the atty resistance)
New Power = (4.7)^2/R

The ratio of new power to old is:

=((4.7)^2/R) / ((3.7)^2/R) - The R's cancel out
= 22.09/13.69 = 1.613 - Remember, this is a ratio, so it does not have any units.

Thus the power is increased by about 61%, which is actually a little higher than the 50% I quoted. This is where I got the 50% number from (I didn't put it in the battery life calculations because I thought 50% was more straighforward). This actually increases the current draw by about 27% rather than 25%, so I guess I was a little bit low in my estimate.

I think watt-hours are quite meaningful in the situation given here. Watt-hours is a measure of the energy stored in the batt. I agree there is a dynamic involved and I think that is why the V^2 calculation fails here. That 3.7 volts of the "old power" is not a constant throughout the discharge life of the batt. Although the regulated voltage output of the eGo, unloaded, is in the neighborhood of 3.45 volts, this is achieved by varying the duty cycle over time. The native voltage that the eGo circuit board "sees" is, I think, the same as a Riva, non-PWM. Generally with most any step up or step down of voltage there will be some energy loss. I am not aware of any methods that would be 100% efficient although there do seem to be some somewhat close to 100%. But, that the standard sized eGo is rated at 650 mah and the Riva at 750 mah is, I think, an indicator of the inefficiency of the conversion from native voltage of the batt to a PWM voltage output. I think, if the heat from the eGo booster is 50% greater than some "old" heat, the hours of use will decline by 50% as well. That does not mean the eGo booster is bad. Just that one can not expect a 50% heat boost and only a 25% or so time loss.

 

[MasterofChaos]

JW50,

Sorry, but you are mistaken. It is not watts that is stored in the battery but charge. You are mixing your units of measurment.

You see, the discharge from the battery (the time the battery will last) is linearly related to the amount of charge that is discharged from the battery. That is why the battery capacity is measured in mAHr, not WHr.

Watts, on the other hand, are quadratically related to the current (and thus the charge). That is why you can get a 50% increase in the "wattage output" (if that phrase even makes sense) for a 25% increase in current output.

Look at it this way: The current is the thing that is being drawn out of the battery (you have to agree with that, right? The battery stores charge), the current generates the power (which is the "watts"), which is quadratically related to the current.

You can't say that the power draw is increased by 50% so the battery life is decreased by 50%. It is not the power that you use to measure the capacity of the battery. Its the current. The battery doesn't really store energy (of course it stores energy, but that's not what you are drawing out of the battery when you discharge it. You are drawing out current).

And you CAN expect a 50% heat boost with a 25% time loss. Its because of the quadratic relationship between current and power.

All you have to do is use the device, and you will see that there is a roughly 50% increase in the vapor output from the atty. And users have made measurements showing a 25% or less decrease in battery life.

Doesn't that pretty much back up the assertions?

 

[JW50]

JW50,

If you look at my Post #24 on this thread, it explains in detail why the booster won't work on a Riva, or any of the unregulated or 3.7 volt fatt-batt devices. It comes down to the fact that the Booster tries to keep the output voltage constant, rather than letting it drop when the battery can't handle the load.

And Vocr, I doubt that the booster would work on any 4.2 volt fat-batt devices unless they had an IMR (high-drain) battery in them. The battery would just not be able to supply the current without a big voltage drop and then you would get that run-away voltage drop thing that I described in that post.

That is pretty standard for any Buck/Boost circuit, BTW. If you look at the Provari (I guess the Darwin doesn't count, because it has its own batteries built in) or any other PVs that use a Buck/boost, they all say you have to use high-drain batteries. Its really not at all unique to the eGo Booster.

Chaos - I indicated previously a didn't understand the explanation at #24. Still don't. Internal resistance should be same for eGo and for Riva. However, what it is looking like to me is that the eGo booster needs to "see" alternating or pulsed voltage or, alternatively, that rms voltage above 3.45 brings about a protection mode of the eGo booster. Again, this is not anti-eGo booster, as it is clearly stated the the booster is not for Riva like batts, but instead a curiosity question. It may be your explanation is right on the mark. But, for me, it is an explanation that does not "penetrate".

 

[GIMike]

Let's make it easy, it's a scheme from the makers of the device, in that it has a built in camera to see what type of battery you're using as to whether or not it will work. Plus it checks to see if you're in the lady's locker room at which point it starts broadcasting video back to the manufacturer which also lowers your battery life. But it helps them with some extra cash on their lady's locker room website. Not sure how that figures into the equation....

 

[JW50]

JW50,

Sorry, but you are mistaken. It is not watts that is stored in the battery but charge. You are mixing your units of measurment.

You see, the discharge from the battery (the time the battery will last) is linearly related to the amount of charge that is discharged from the battery. That is why the battery capacity is measured in mAHr, not WHr.

Watts, on the other hand, are quadratically related to the current (and thus the charge). That is why you can get a 50% increase in the "wattage output" (if that phrase even makes sense) for a 25% increase in current output.

Look at it this way: The current is the thing that is being drawn out of the battery (you have to agree with that, right? The battery stores charge), the current generates the power (which is the "watts"), which is quadratically related to the current.

You can't say that the power draw is increased by 50% so the battery life is decreased by 50%. It is not the power that you use to measure the capacity of the battery. Its the current. The battery doesn't really store energy (of course it stores energy, but that's not what you are drawing out of the battery when you discharge it. You are drawing out current).

And you CAN expect a 50% heat boost with a 25% time loss. Its because of the quadratic relationship between current and power.

All you have to do is use the device, and you will see that there is a roughly 50% increase in the vapor output from the atty. And users have made measurements showing a 25% or less decrease in battery life.

Doesn't that pretty much back up the assertions?

I think a battery is better defined as an energy storage device instead of a charge storage device. This from Google summary of Wikipedia on batteries: "An electrical battery is one or more electrochemical cells that convert stored chemical energy into electrical energy." Watt-hours is energy. But I do agree that it is possible to obtain 50% more vapor with only a 25% loss of time. But, lots of other factors other than just current or just watts is involved there.

 

[JW50]

Let's make it easy, it's a scheme from the makers of the device, in that it has a built in camera to see what type of battery you're using as to whether or not it will work. Plus it checks to see if you're in the lady's locker room at which point it starts broadcasting video back to the manufacturer which also lowers your battery life. But it helps them with some extra cash on their lady's locker room website. Not sure how that figures into the equation....

An interesting way to look at it. To a male it may seem like a 50% boost but to a female it may only seem like 25%. The duty cycle needs to be accounted for to get to which is correct.

 

[JW50]

On the issue of mAh and watt-hours, here is a discuss that occurred in 2008 on another forum (dealing with laptops):
By a Stukpixel:

I was looking at an external battery, but after reading into it, I realized that there is something mathematically wrong with its designate capacity.

20000 mAh/74wh .

(watt hours = mA * voltage)
Now assuming wh means the standard watt hours:
74wh /= 20 * any voltage above 4!


can someone explain why I'm wrong, or confirm my logic?

Then from a JPZ:

Wow...

Ok, a LOT needs to be cleared up here. First, some key terms:

energy - I don't have a good definition, but you all should know what it is. One unit of measurement for energy is the joule.
charge - The actual charged particles that transfer energy. A coulomb is a standard unit of charge.
voltage - potential difference, the energy of a charged particle. Measured in volts(joules per coulomb).
current - how much charge is flowing. Measured in amps(coulombs per second).
mAh - milli-amp-hours. Another unit of charge commonly used in electronics industry. This is a measure of how many charges a battery can store for later use.
power - Rate of energy transfer(joules per second). Measured in watts.

Now a few more equations:

Power = Current * Voltage (finding power for circuits)
Energy = Power * Time

There is a huge misunderstanding in this thread, and that is the difference between watts and watt-hours. Watts is a rating for power, and watt-hours is for energy. Remember I said that power is rate of energy? Rearrange this and you get Energy = Power * Time. This is where watt-hours comes from.

Ok, now to address the issue here. This battery is clearly capable of being used with multiple devices and can output different voltages. Therefore the battery must have some sort of device between the battery to increase the voltage from the battery depending on the device. My point is, the battery itself likely does not charge to one of the listed output voltages.

With a little math, a 74Whr battery with a charge capacity of 20Ah would have an output voltage of 3.7 volts. A very common voltage for your standard lithium cell. And this battery is lithium based.

Now that I do the math and see that I have gotten some similar numbers as some of you(including the OP, for one), I see you seem for the most part to know what you are doing and there were just a few typo's throughout your posts. But I'll leave all the information above in case anyone else finds it useful.

The bottom line: It is probably safe to trust that the battery is a 74Whr battery. Go by the energy rating of the battery. That's why it is there. Charge capacities(mAh or Ah) are usually just for marketing ploys and really don't matter.

Not saying JPZ is correct. But the mah vs watt-hour issue has caused some confusion in the past.

 

[GIMike]

On the issue of mAh and watt-hours, here is a discuss that occurred in 2008 on another forum (dealing with laptops):

20,000 mah??? Who's making the first laptop battery mod? Just make sure it's got a PWM circuit and limited to 3.7v so I can use my ego booster on it

 

[JW50]

20,000 mah??? Who's making the first laptop battery mod? Just make sure it's got a PWM circuit and limited to 3.7v so I can use my ego booster on it

Always an interesting perspective! But actually, a pass through would probably work. Just leave computer in sleep mode to minimize its draw on those 20,000 mah's or 74 wh's.

 

[GIMike]

But with a 20,000 mah battery I wouldn't have lug around the rest of the weight of the laptop with me making it more portable than a whole laptop, and it should last what, a whole month between charges? Ok, yeah, that is kinda overkill lol

 

[JW50]

But with a 20,000 mah battery I wouldn't have lug around the rest of the weight of the laptop with me making it more portable than a whole laptop, and it should last what, a whole month between charges? Ok, yeah, that is kinda overkill lol

I guess even if carrying that 20,000 mAh batt causes you to walk with limp or makes you to sag on one side - beats lung cancer. Then again, we don't fully understand the long term consequences limping or sagging so maybe best to await FDA approval. Oops. I see a problem in the horizon. Laptops as drug delivery devices.

 

[JW50]

Some comments concerning the calculations at post #42. If one looks at the oscilloscope readings here Artisan's Workshop - Ego Booster.m4v - YouTube
what is shown is that the voltage of the eGo without the eGo booster pulses from zero volts to 3.7 volts. This is a loaded condition. The duty cycle, at least as best as I can find on the output readings of the scope, is not shown. Now when the eGo booster is added to the eGo and the booster is dialed to its highest output, the reading of the scope shows pulsed output of zero to 4.7 volts. Again, I can not find duty cycle. However, neither the 3.7 volts or the 4.7 volts is a continuous read, something the formula at post #42 is assuming. Considering the duty cycle the eGo pulse. On the voltmeter it read 3.45 in the video in an unloaded state. But in actuality this 3.45 was an average of some maximum voltage for certain periods of time and zero for certain periods of time. From testing I have done on a Riva, this maxium voltage was probably around 4.22 volts. That is, unloaded the duty cycle of the eGo is around 0.815. (i.e. 3.45 divided by 4.22). Loaded - we don't know but from testing I have done on loaded Riva's and loaded eGo's I would say it is in range of 0.848. If this is case, then the "old power" of the formula of post #42 should be (3.7*0.848)^2/R. There would be a similar correction for the maximum 4.7 voltage as well. Its duty cycle is likely something quite different from 0.848. I suspect with these corrections one will get to a result that when watts increase by 50%, time will decline by 50%. This, in my opinion, does not mean that the voltage boost of the eGo booster is not there - simply that it will come with a price of shortened battery time on a single charge.

 

[MasterofChaos]

JW50,

Oh, sorry, I didn't see that post. Really, it can be simplified as follows:

Without the eGo Booster, the battery voltage can drop which drops the current.

With the eGo Booster, when the battery voltage drops, the current goes up.

The reason it works on an eGo rather than a Riva is because the eGo sits at a lower operating point to begin with, so it doesn't drop with the eGo Booster on it. That is kind of an oversimplified explaination, however you can look at it as the eGo having more "head room" built into it because the output voltage is set lower, however the Riva is designed to operate close to its upper limits right out of the box.

That might explain why Rivas seem to be more prone to dying as well, aside from manufacturing issues.

 

[MasterofChaos]

JW50,

Your duty cycle estimates are pretty close; I've meausured the eGo's loaded duty cycle directly and it sits at about 75-80% with or without the booster.

Yes, you have to put the duty cycle in there to get the absolute value of the power, however since the duty cycle is the same whether its 3.7 or 4.7 volts (remember I was calculating power INCREASE not the absolute power) it cancels out. I've left duty cycle off some of the calculations since I figured it made it even more confusing, and its pretty constant whether the booster is on there or not.

As far as comparing a Boosted eGo to a Riva, I really can't say. I haven't done a lot of measurements with Rivas, since the ones I did have tended to blow up pretty easy. Remember, no claims were ever made comparing a boosted eGo to a Riva, all that was claimed was that it made the eGo better.

If someone has Rivas and an eGo Booster to test, that would definitely be an interesting comparison. I can't really say what a side-by-side comparison would be like. Part of the problem with that is the atty/carto you are using, which changes the experience quite a bit, and adds a lot of variables to the test.

 

[JW50]

JW50,

Your duty cycle estimates are pretty close; I've meausured the eGo's loaded duty cycle directly and it sits at about 75-80% with or without the booster.

Yes, you have to put the duty cycle in there to get the absolute value of the power, however since the duty cycle is the same whether its 3.7 or 4.7 volts (remember I was calculating power INCREASE not the absolute power) it cancels out. I've left duty cycle off some of the calculations since I figured it made it even more confusing, and its pretty constant whether the booster is on there or not.

As far as comparing a Boosted eGo to a Riva, I really can't say. I haven't done a lot of measurements with Rivas, since the ones I did have tended to blow up pretty easy. Remember, no claims were ever made comparing a boosted eGo to a Riva, all that was claimed was that it made the eGo better.

If someone has Rivas and an eGo Booster to test, that would definitely be an interesting comparison. I can't really say what a side-by-side comparison would be like. Part of the problem with that is the atty/carto you are using, which changes the experience quite a bit, and adds a lot of variables to the test.

I see a little concession here in that you agree that duty cycle is part of equation. However, contrary to your belief that it cancels out - I'm not so sure. If you go here Boost converter - Wikipedia, the free encyclopedia you will find an equation for a boost circuit of D= 1 - Vi/Vo. If Vi is 3.7 and Vo is 4.7 then D=0.2128. So, how do you factor that in?

Lets go backwards. Here is what you said:

New Power = 150% of Old Power = 1.5*Old Power
or
Inew^2*R = 1.5*Iold^2*R.

The Rs cancel out, leaving:

Inew^2 = 1.5*Iold^2

If we take the square root of each side of the equation, we get:

Inew = 1.224*Iold

Thus the new current is 122.4% of the old current.

OK. But now let's do it with voltage. It would go as follows:

New Power = 150% of Old Power = 1.5*Old Power
or
Vnew^2/R = 1.5*Vold^2/R.

The Rs cancel out, leaving:

Vnew^2 = 1.5*Vold^2

If we take the square root of each side of the equation, we get:

Vnew = 1.224*Vold

Thus the new voltage is 122.4% of the old voltage.

Let's presume we agree that P=IV (Power=current times voltage). So what we have is if current increases by 22.4% AND voltage increases by 22.4% we have 50% more power than before.

Then you say:

What this means is if you increase the current to the atty by 22.4%, power dissipated by the atty increases by 50%. Also, if you increase the current into the atty by 22.4%, then you will decrease the battery life by the exact same amount. So a fully Boosted battery’s life will be 77.6% of a non-Boosted battery’s life. Not a bad trade off considering the significant improvement in vapour and flavour provided by the Ego Booster.

This I would edit as follows:

What this means is if you increase the current to the atty by 22.4% and voltage to the atty by 22.4% , power dissipated by the atty increases by 50%. Also, if you increase the current into the atty by 22.4% and increase the voltage to the atty by 22.4%, then you will decrease the battery life by the exact same amount of 22.4% times 22.4% or ~50%. So a fully Boosted battery’s life will be 50% of a non-Boosted battery’s life. Not a bad trade off considering the significant improvement in vapour and flavour provided by the Ego Booster.

 

[MasterofChaos]

JW50,

As far as the duty cycle thing is concerned, with your Wikipedia Booster reference you are confusing the pulsing that is created by the booster circuit (which is operating at 2.4 MHz, and is filtered out anyway, so it is irrelevant) with the pulsing that is done by the eGO Battery (that is around 100 Hz). The pulsing you see on the video is the PWM of the eGo Battery, and has nothing to do with the Booster. It is there whether the Booster is there or not. That is why the duty cycle cancels out.

Look, if you still doubt the 50% power increase then okay, it seems like no amount of logic, calculations, first-hand testimonials by users or measurements will convince you.

I stand by my statements and calculations. If you want to argue an alternative go right ahead, however I would suggest you wait until you test the unit yourself before you cast aspersions what its doing.

And please don't edit my statements, since you are now attributing something to me that is incorrect. Even if what you are trying to assert is true (which its not), 22.4% times 22.4% is not 50%. It would be 5%. So even on a basic mathematical level your numbers don't make any sense.

 

[cozzicon]

On a lighter note:

If you give me an oscillator, an iron core, about 200 feet of fine copper wire, and 1.5 volts DC: I can show you really high voltage vaping!!