PWM and lavatube effecting taste?

[quisp65]

I bought a $20 volt/ohm meter from Sears. It's not one that screws on if that's what your asking. I don't know the voltage under load. My LT said it was 2.6 ohms. But it was saying everything was 2.6 ohms till I put a few atomizers I found them rated 1.7. I was trusting it over the LT. Maybe the LT is more accurate. If thats the case then me not ever going over 3.4 volts would make more sense. I'll try the flavor on an atomizer and see how that effects it...working today though..

Also... if I had a lavatube from Apolloecig that's the original I should be able to get the same type of voltage as my Kgo and not worry about all this? I can get a tube only for $40. I like that better than having to shop different devices for each unit. But that's something I would wait a while on. Still gotta put this thing through it's paces and see if my attitude changes or it changes my vaping style.

 

[Blitzer]

I bought a $20 volt/ohm meter from Sears. It's not one that screws on if that's what your asking. I don't know the voltage under load. My LT said it was 2.6 ohms. But it was saying everything was 2.6 ohms till I put a few atomizers I found them rated 1.7. I was trusting it over the LT. Maybe the LT is more accurate. If thats the case then me not ever going over 3.4 volts would make more sense.

Also... if I had a lavatube from Apolloecig that's the original I should be able to get the same type of voltage as my Kgo and not worry about all this? I can get a tube only for $40. I like that better than having to shop different devices for each unit. But that's something I would wait a while on. Still gotta put this thing through it's paces and see if my attitude changes or it changes my vaping style.

The resistance checker on the new lavatube is pretty good.

The V1 lavatube will perform just like your KGO. Except the KGO voltage will slowly go down as you use it. The V1 lavatube will hold the voltage you set, until the battery can't keep up anymore. With your ohm range, and voltage range- you should have NO trouble with the V1 Lavatube.

 

[quisp65]

Just to update if anyone is following this thread...

LordDavon gave a simplified rough estimate of what PWM did with the V8...

"Roughly, the PWM device is 25% more powerful than the constant voltage device"

Hopefully I'm not taking him out of context..... but given that the lavatubes that use PWM go down to 3.0 volts this would give a lavatube that uses PWM set at 3.0 volts the feel of something running at 3.75 volts. Being that ego devices usually run around the 3.5V mark on voltages, a lot of the carto market is tailored for devices to taste good at this voltage. This would make the PWM lavatubes be at the upper limit for many cartos on the market.

We know the lavatube 1.5s have both pros and cons to their l-ryder brothers, and the 1.5s are very frequently mentioned as allowing a larger range of dual coil devices to be placed on them. It would seem it should also be mentioned along side this just as much, something similar to my critique above.

Anyway.... I hate sounding negative.... I write this only for the purpose of informing or learning about the lavatube. I recently got the 2.5v head inside my Vivi Nova and with a head measuring 2.8 ohms I can run my favorite flavor at 3.0Volts and it tastes great. The Vivi looks good on this thing and I think the long wick newer version are a definite improvement over the two previous heads before it.

 

[Blitzer]

Lord Davon was talking about the v8. The lavatube with PWM won't go to 7.2 volts.
I've looked at it on a scope. The max you can get is 6.2 volts unloaded out of mine.
When you put an atty on the lavatube, the peak voltage drops based on the resistance.
Our lavatubes are a little stronger than an analog power source would be, but not 25% more for sure!
There is a thread on my lavatube scope findings if you care to look.

 

[quisp65]

Ah... ok... thanks for clarifying... this is all way over my head for me to probably to grasp real well... but I'm sure you already knew that

How much extra percentage of power would you rate the PWM lavatube over actual volts displayed? I know this might be a guess and probably wouldn't be consistent depending on circumstances. But rough average maybe?

ty for all your knowledge and discussion, I hope you have been continuing in this thread because you enjoy talking about the lavatube and not because of the need to correct a newb giving wrong info about it I would feel guilty if it was the later

 

[Blitzer]

First, I don't know if you are a noob! I believe you when you say your LT is putting out more power
than you feel like it should. Mine is too in some cases. I will try to explain.

Your question is not that simple to answer, without getting into a bunch of ohms law.
For instance: our lavatubes will set the peak voltage based on the ohms of the atty.
In Lord Davon's example he used peak of 7.2v, 2 ohms, and 40% duty.
The V8 uses constant peak voltage, and delivers average voltage by pwm.
The LT has the PWM, but the peak voltage is set by the atty resistance.
If I put a 2 ohm carto on my lavatube, it will put out a max. amplitude pulse of 4.35 volts.

OK, if you had an analog device, set to 4.7 volts on a 2 ohm carto, it would be EXACTLY the same
as the lavatube with a 2 ohm carto, and set to 6v on the display.
(this is 100% duty cycle, on all the time at the peak voltage for the 2 ohm carto)
In this case, the pwm "increase" in power is 0. Maybe you could also say 4.35 volts is 27% below 6v.

In another case, we put 3v on the lavatube display, and this gives 50% duty cycle.
Since our carto is still 2 ohms, our peak volts is still 4.35 volts. Using lord davons formulas i came up with a 22.% increase
from an analog device. The answer to your question will be different based on your carto atty.

There is an easier way to deal with this. Imagine you are a 2 ohm atomizer getting 3 volts all the time.
You make a certain amount of vapor. Now imaging you are the same atomizer getting a voltage that turns off and on. When the voltage is on, you feel 4.35 volts. Wow! I am getting hot! But then it turns off. Oh, cooling down now. I Mr. Atomizer know the voltage across me is averaging the same or less, but for those times it's 4.35 v, I am getting fried!

Look into Ohms law formulas and google PWM for more info. The relevant formulas are:

1. Duty cycle = pulse repetition frequency * Pulse width
2. Average power = Peak power * duty cycle
3. Peak power = average power / duty cycle.

All the other ohms law equations are still relevant for calculation of voltage, resistance, power and current during the instantaneous sample time.

I wish I could tell you, yeah pwm is always 10, 15, 20, 30, 40% more than an analog device, but it's more complicated than that.

 

[Blitzer]

First, I don't know if you are a noob! I believe you when you say your LT is putting out more power
than you feel like it should. Mine is too in some cases. I will try to explain.

Your question is not that simple to answer, without getting into a bunch of ohms law.
For instance: our lavatubes will set the peak voltage based on the ohms of the atty.
In Lord Davon's example he used peak of 7.2v, 2 ohms, and 40% duty.
The V8 uses constant peak voltage, and delivers average voltage by pwm.
The LT has the PWM, but the peak voltage is set by the atty resistance.
If I put a 2 ohm carto on my lavatube, it will put out a max. amplitude pulse of 4.35 volts.

OK, if you had an analog device, set to 4.7 volts on a 2 ohm carto, it would be EXACTLY the same
as the lavatube with a 2 ohm carto, and set to 6v on the display.
(this is 100% duty cycle, on all the time at the peak voltage for the 2 ohm carto)
In this case, the pwm "increase" in power is 0. Maybe you could also say 4.35 volts is 27% below 6v.

In another case, we put 3v on the lavatube display, and this gives 50% duty cycle.
Since our carto is still 2 ohms, our peak volts is still 4.35 volts. Using lord davons formulas i came up with a 22.% increase
from an analog device. The answer to your question will be different based on your carto atty.

There is an easier way to deal with this. Imagine you are a 2 ohm atomizer getting 3 volts all the time.
You make a certain amount of vapor. Now imaging you are the same atomizer getting a voltage that turns off and on. When the voltage is on, you feel 4.35 volts. Wow! I am getting hot! But then it turns off. Oh, cooling down now. I Mr. Atomizer know the voltage across me is averaging the same or less, but for those times it's 4.35 v, I am getting fried!

Look into Ohms law formulas and google PWM for more info. The relevant formulas are:

1. Duty cycle = pulse repetition frequency * Pulse width
2. Average power = Peak power * duty cycle
3. Peak power = average power / duty cycle.

All the other ohms law equations are still relevant for calculation of voltage, resistance, power and current during the instantaneous sample time.

I wish I could tell you, yeah pwm is always 10, 15, 20, 30, 40% more than an analog device, but it's more complicated than that.

 

[Blitzer]

sorry about the double post, the stupid data entry box froze up on me!

 

[Blitzer]

In the world of radar, we do exactly this kind of stuff all the time.
You can build a radar to handle 4000 watts of average power fairly cheaply,
but once you look at the peak power things get a lot more expensive.

I used to work at a place that tracked stuff in space with 5 megawatts pulses.
We had to be concerned with the average power, and the peak power at the same time.

Motor control circuits and light dimmers use pwm for similar reasons to our ecig. The engineer has to
make sure his peak power doen't fry the device he is controlling. PWM allows much better control of lights and motors than dc does. You can use pwm to have a motor barely moving, or have the motor screaming fast. PWM is a good thing.

 

[Blitzer]

One final thought. If money was no object, you could design a PWM power supply
to handle .001 ohm cartos from 0- nearly infinity. Weight would also be a problem, you couldn't carry it.
The original Lavatube couldn't keep up with the current demands of low res cartos.
PWM is a more efficient way to control things than dc voltage is.
The new lavatube does much better with low res cartos, thanks to PWM but could use more power.
Maybe the new version will address this.
The engineers who designed this did a fantastic job, increasing the performance of the original LT, without
increasing the size.

 

[videodave]

quisp65,
I just ordered my new LavaTube (and, of course, am stalking the mail carrier waiting for it) but my favorite combo right now is a KGO with 1.5 ohm dual coil cartos. I am told this is probably the WORST thing you can vape on a LT. So I am biting the bullet and also ordered some 3.0 ohm single coil cartos for the LT. I figure this way I will be able to really make use of the variable voltage of the LT and search for my "sweet spot." Some day I might find a device that works well with both the KGO and the LT. But in the mean time I am going to switch devices between the two. I hope the experience will be worth the extra effort. -videodave

 

[VapingTurtle]

... Now imaging you are the same atomizer getting a voltage that turns off and on. When the voltage is on, you feel 4.35 volts. Wow! I am getting hot! But then it turns off. Oh, cooling down now. I Mr. Atomizer know the voltage across me is averaging the same or less, but for those times it's 4.35 v, I am getting fried!
...

Where this analogy falls WAY short, as I'm sure you know, Blitzer, is that the heating element is not a digital device that heats up and cools down anywhere near instantaneously, nor anywhere near the switching frequency of the PWM circuit, just as a motor on a PWM power source does not stop and start with each pulse.

 

[Blitzer]

Where this analogy falls WAY short, as I'm sure you know, Blitzer, is that the heating element is not a digital device that heats up and cools down anywhere near instantaneously, nor anywhere near the switching frequency of the PWM circuit, just as a motor on a PWM power source does not stop and start with each pulse.

Yes, you are correct this was an over simplification! It does help people who don't want math to understand, I think.

The correct way to look at this is to calculate vrms, which can give us average power. Yes, vrms is for ac voltage
but it also is valid for pulse trains. More accurate in fact, than squaring the volts,then divide by ohms, then divide by duty cycle.

Our lavatubes set a peak voltage based on the atty ohms we have. (to keep us from burning up something)
Lets see what a 2 ohm atty gives us. I will use 10w as the max power of our lavatube. They are advertized
as 12w max on some sites. I measure mine closer to 9.5 watts. We will use 10w for simplicity.
We will use the formula E=sqrt(P*R). plugging in power and ohms:E=sqrt(10*2). answer: E=4.47v

We have our peak voltage for this 2 ohm atty. (4.47v) This voltage doesn't change. Only the pulsewidth changes.
I have verified this on a scope under load.

We have selected 3volts on our lavatube. (3v/6v=.5 duty cycle)
We need to calculate vrms, before we can calculate power.
first we square our peak voltage: 4.47 * 4.47 = 19.98
THen we get the average of this value by multplying it by our duty cycle:
19.98 *.5= 9.99
Then we take the sq. root of this: sqrt(9.99)= 3.16vrms

NOW we can calculate our power: P=E^2 / R (3.16*3.16)/2= 4.99w average power

The results of this same 2 ohm atty, across the entire range of our lavatube:

LT display= 3 volts VRMS= 3.16v Avg. Power=4.99w
LT display= 4 volts VRMS= 3.63v Avg. Power=6.58w
LT display= 5 volts VRMS= 4.07v Avg. Power=8.28w
LT display= 6 volts vrms= 4.47v Avg. Power=10w

One final thing we can do is compare our 2ohms atty on our lavatube to a voltage source (lets call it batt pwr),
capable of driving as much current as we want:

3v LT pwr= 4.99w batt pwr= 4.5w
4v LT pwr= 6.58w batt pwr= 8w
5v LT pwr= 8.28w batt pwr= 12.5w
6v LT pwr= 10w batt pwr= 18w

You can see the lavatube power is a little stronger at the bottom 3v setting (10%)
The LT voltage & power rapidly fall behind the batt pwr voltage source.

 

[Blitzer]

In case somebody reading the above doesn't know, VRMS means "voltage root mean squared". It's the accurate way of expressing the dc equivalent voltage of a sinusoidal voltage, or of a pulse train. Google "square wave rms", and you will find a ton of info. A lot of wrong info, and a lot of right info! Lots of folks don't understand it! Still, it's really the most accurate way to talk about the dc equivalent voltage of our lavatubes under load.

 

[billo]

RMS voltage is what causes heating, NOT average voltage. It's very simple:

Average voltage= peak voltage * duty cycle
RMS voltage = peak voltage * square root of duty cycle

So, for example, if you had six volts that is on half the time and off half the time the duty cycle would be .5

So, average voltage = 6V * .5= 3V

Rms voltage= 6V * square root of (.5)= 6 V * .707 = 4.24 v

The 4.24 volts is what causes the heating. So it would be like putting 4.24 volts DC--a 4.24 V battery-- across the heating element.

However, if you measure the average pulsed voltage you would get 3 volts, which is not the value that determines the amount of heating.

Note, that for pulsed or AC voltage, RMS is always higher than average, so the heating
effect will always be higher than you think by reading the average voltage. An extreme example is your house wiring. It has an RMS value of about 115 Volts, but ZERO average voltage !

Another confusing factor is what value would you read on a DC digital voltmeter set to read voltage ? I'm not sure, and it might even depend on the type of meter you have.

A sure way would be to have an oscilloscope and measure the peak value and duty cycle then use the above formula. Another way would be to use a voltmeter that measure RMS voltage, but these tend to be expensive and not readily available.

The above could account for statements like "my brand x hits much harder at 4 volts than my brand Y." Yes, if brand x if pulse-width modulated and brand Y is not, and you're measuring average voltage of each, then brand X will have an RMS
voltage that is higher and will do more heating.

PS, I am an Electrical Engineer.

 

[Blitzer]

RMS voltage is what causes heating, NOT average voltage. It's very simple:

Average voltage= peak voltage * duty cycle
RMS voltage = peak voltage * square root of duty cycle

So, for example, if you had six volts that is on half the time and off half the time the duty cycle would be .5

So, average voltage = 6V * .5= 3V

Rms voltage= 6V * square root of (.5)= 6 V * .707 = 4.24 v

The 4.24 volts is what causes the heating. So it would be like putting 4.24 volts DC--a 4.24 V battery-- across the heating element.

However, if you measure the average pulsed voltage you would get 3 volts, which is not the value that determines the amount of heating.

Note, that for pulsed or AC voltage, RMS is always higher than average, so the heating
effect will always be higher than you think by reading the average voltage. An extreme example is your house wiring. It has an RMS value of about 115 Volts, but ZERO average voltage !

Another confusing factor is what value would you read on a DC digital voltmeter set to read voltage ? I'm not sure, and it might even depend on the type of meter you have.

A sure way would be to have an oscilloscope and measure the peak value and duty cycle then use the above formula. Another way would be to use a voltmeter that measure RMS voltage, but these tend to be expensive and not readily available.

The above could account for statements like "my brand x hits much harder at 4 volts than my brand Y." Yes, if brand x if pulse-width modulated and brand Y is not, and you're measuring average voltage of each, then brand X will have an RMS
voltage that is higher and will do more heating.

PS, I am an Electrical Engineer.

I agree that RMS is the heating equivalent voltage.
I can't see anything different in this from what I have said before, except my numbers were based on
oscilloscope measured peak voltage and duty cycles.

 

[Blitzer]

Actually Billo, I like your method of calculating VRMS better! Gives the same answer. As in so much of electronics,
there are several algorithms that will achieve the same results. Your formula of getting the sq. root of the duty cycle,
and multiplying by peak voltage works just the same, and is quicker. Thanks.

 

[billo]

Blitzer..

Yes, your explanation is correct too.

Edit: Weird coincidence that we both reposted at the same time !

 

[jonnypanic]

Glad I found this thread as I was getting concerned about my new purchase.
Yesterday I received a Vector Variable Volt and was getting concerned because I could not get a good vape. Sweet spot with current juice is 3.6V at 3.2ohms, which according to Ohm's Law = 1.1 amps and 4 watts. Yesterday I had a 1.8 ohm head (reading 2 ohms) and it tasted burnt at 3 volts.

 

[Vaptor]

My lavatube set at 3V puts out nearly 4.5V. Voltages get closer to the real Voltage the higher you set it. But even than as soon as you hit the button (if it doesn't actually misfire as it happens) it supplies the power to the coil bang on so it's hot almost immediately and that also affects taste. Good mods bring up the initial Voltage slower. 510 manual batteries are good example. It's designed that way, not because the battery is small. I made a mod out of 510 PCB and use 18650 in it. It still kicks in slow but gives a nice smooth vape.