Battery Voltages -- Surprise!

[Rocketman]

I'm going to outline a test jig someone can make at home to play around with measuring PWM e-cig voltages.
This is something I made as a show-and-tell many years ago to demonstrate the effects of AC and DC power.
Now days, accurate RMS, and pulse power digital instruments can be bought, but back in the day, AC voltage and power equipment was calibrated using a 'Thermal Transfer Standard'. The model I used was a Holt TV-1, but this little home made device I'm going to describe can be surprisingly accurate.

A transfer standard performs a comparison to a known, or easily measured value (like DC volts) using a fundamental and defined property of the transfer device to measure the unknown, and usually difficult to measure value (like AC volts).

RMS volts produce the same 'Heating effect' as DC volts. So, why not put the AC volts to a resistive heater and measure the temperature, then swap to DC volts and adjust for the same temperature? Then measure the DC volts.

But, the show-and-tell mockup used a automotive 12 volt turn signal bulb (not a halogen cycle bulb) and a silicon photo-voltaic cell (in a box). The lamp filament will not respond in a linear manner but X volts AC will produce the same heating (and light output) as X volts DC.
Put a cartomizer connector on it and hook up an e-cig. Measure photo cell output. Hook up a variable voltage DC power supply and adjust for the same light output.

The ONLY device that needs to be calibrated is the DC voltage measurement. Basic function of the photo cell, the lamp, wiring, provide "sameness" for transferring the measurement to determine the AC RMS voltage.
A Therm-optic Transfer Volt Meter.

 

[Stonemull]

average power is identical to the RMS power, it the sum of all powers over time, in which case you need the true voltage to calculate the square to give you power.
you cannot derive average power however by using the average voltage, that is the whole point of the post. The sum of squares of the instantaneous voltage is not the same as the mean of the voltages.
In the simple 50% duty cycle example above, the correct result for average power would be if you measured the peak power and divided by 2 .. 8W, the RMS power is 8W as well.
totally agree with the light bulb and of course its a resistive load, true RMS meters used light bulbs and a photoelectric cell for years or an analog multiplier cct, these had limited bandwidth though.

 

[Stonemull]

you could make that or simply measure the peak voltage and duty cycle on a basement quality CRO and the power is = (Vp^2 * dc)/ R where dc is obviously duty cycle. Vp is peak voltage and will with all PWM mods be roughly 100mV or so below cell voltage.

 

[keydcuk]

and the difference between the average power and RMS power for a square wave driving a resistive load is?

(at eGo frequencies the Coil is resistive)

You could drive a incandescent lamp and measure the output, then compare it to DC.
Sounds tough, but it's not.

Speaking from an electronics background, there is no such thing as RMS power. I always get a kick out of manufacturers that mention RMS. It doesn't exist!, only average power. RMS voltages yes, no RMS power.

 

[JW50]

Speaking from an electronics background, there is no such thing as RMS power. I always get a kick out of manufacturers that mention RMS. It doesn't exist!, only average power. RMS voltages yes, no RMS power.

In some ways RMS voltage doesn't exist either. Instead, it's a mathematical result. RMS voltage is that pure DC voltage that produces the same power as a varying voltage signal over a cycle of the voltage variance. Meters exist that output a RMS voltage but such meters are in a manner "calculating" to create an output. The Rocketman mentioned jig, if I understand it, is "calculating" by varying the pure DC input to produce the same temperature/light that was produced by the varying input. Instantaneous power of a pure DC input is a constant number whereas the instantaneous power of a PWM input varies from zero (at those times when there is zero input, duty "off") to a maximum (at times when duty is "on"). The RMS voltage is that number that produces the same sum of instantaneous "powers" over a cycle of varying input as would be produced by that constant DC voltage over the same cycle.

 

[Rocketman]

Speaking from an electronics background, there is no such thing as RMS power. I always get a kick out of manufacturers that mention RMS. It doesn't exist!, only average power. RMS voltages yes, no RMS power.

Sorry to offend anyone with the slip "RMS Power".
We don't RMS power values with respect to time to determine RMS Power.

If we go back to the OP, and the original discussion, it seems the problem everyone is trying to fix is measuring the Joye 510 and eGo output voltage, and the resulting power (in average watts) as delivered to an atty/cartomizer. Those that don't even have an eGo have repeatedly pointed out that a simple, cheap DMM can not provide a meaningful voltage measurement of the PWM output of the Joye devices without using expensive measuring equipment.
RMS measurements of voltage and average measurement for sinusoidal waveforms have been sited as the problem for a simple integrating DMM with 3 to 10 measurements/sec. It is obvious from reading this that everyone needs a Fluke handheld DMM with RMS indicating voltage ranges

As far as I know the PWM output of the Joye products are switched on and off at a somewhat low frequency. Just what is the difference in output power for a square wave (even one with a varying duty cycle) when the frequency is several times the sample rate of the measuring device, when the voltage is measured with an "averaging device" or a "RMS indicating device?

What is the average power over the duration of a typical vape puff delivered to the atty for measurements measured by a DC digital Voltmeter vs that measured by a RMS indicating voltmeter? That is why we are at over 500 posts in this thread.

If I misspelled anything, I hope someone with a Language Arts background corrects me

 

[JW50]

you could make that or simply measure the peak voltage and duty cycle on a basement quality CRO and the power is = (Vp^2 * dc)/ R where dc is obviously duty cycle. Vp is peak voltage and will with all PWM mods be roughly 100mV or so below cell voltage.

In preceding post to above post you indicate that one can not derive average power by using average voltage. However, consider that for PWM that Va=Vp*dc (where Va = average voltage, Vp = peak voltage and dc = duty cycle). Rearranging this is Vp=Va/dc. Then substituting the Vp equivalent in your equation above the power becomes (Va^2/dc^2*dc)/R or Va^2/(dc*R). That is, average power can be derived from average voltage. Then, with last equation, the Vrms (where Vrms is the RMS voltage) is Va divided by square root of dc (i.e. that pure DC voltage that will give an average power of Va^2/(dc*R). In other words, (Va/dc^.5)^2/R equals average power.).

 

[Stosh]

In the 60's for going to the moon, NASA spent millions on development of a zero gravity ball point pen for the missions, the Russians used a pencil.

I use the lowest cost, simplest low frequency PWM I could find, my thumb. If the power is burning the juice I let go of the button, needs to be hotter - hold the button longer. Voltage is not important at all, pulse the button to produce good vapor at a temp that suite you.

Proof of concept : I've vaped a 1.7 ohm atty on a 5v / 3amp supply, WOW 14.7 watts.....not really pulse the button and the juice don't burn, the atty doesn't pop, actual wattage likely about 7-8 watts. YMMV

 

[JW50]

...

What is the average power over the duration of a typical vape puff delivered to the atty for measurements measured by a DC digital Voltmeter vs that measured by a RMS indicating voltmeter? That is why we are at over 500 posts in this thread.

...

I think average power would be Vrms squared divided by resistance if the voltage is RMS voltage (i.e. measured with RMS voltage meter) or it is Vdc (voltage on dc scale of ordinary volt meter) squared divided by resistance divided by duty rate. Note if duty rate is 1.0 that average power equals Vdc squared divided by resistance and that Vrms is same as Vdc.

 

[Rocketman]

When you hook a voltmeter, 4 dollar DMM, on the 20 volt DC range to an eGo driving an atty:
What does the meter read? I mean the display reading, not multiplied by duty cycle, not squared then square rooted ,
the actual meter reading?

Does it read 4.1 volts? 2 volts? 3.4 volts?
( I know what mine reads)
Just what math do you need to do to the cheap DMM reading to mathematically correct it?
Here's a shot of an ego, a carto, and super cheap DMM.
Could someone come up with an equation to mathematically correct this display reading (ignoring meter calibration inaccuracy) so I could calculate average power being delivered to the carto (3 ohm).
This meter obviously does not indicate RMS voltage. So it must be wrong. But, it is sort of square

BTW, what is the 'Crest Factor' of the eGo output into a resistive load?

 

[Rocketman]

Proof of concept : I've vaped a 1.7 ohm atty on a 5v / 3amp supply, WOW 14.7 watts.....not really pulse the button and the juice don't burn, the atty doesn't pop, actual wattage likely about 7-8 watts. YMMV

Right thumb, or left thumb

 

[JW50]

In the 60's for going to the moon, NASA spent millions on development of a zero gravity ball point pen for the missions, the Russians used a pencil.

I use the lowest cost, simplest low frequency PWM I could find, my thumb. If the power is burning the juice I let go of the button, needs to be hotter - hold the button longer. Voltage is not important at all, pulse the button to produce good vapor at a temp that suite you.

Proof of concept : I've vaped a 1.7 ohm atty on a 5v / 3amp supply, WOW 14.7 watts.....not really pulse the button and the juice don't burn, the atty doesn't pop, actual wattage likely about 7-8 watts. YMMV

I suppose one might call this method of vaping - vaping by the seat of your pants. Nothing at all wrong with it. But at three amps, best to have a high quality button to pulse with and batteries that can withstand some high momentary drains. Suspect there are some added risks there just like flying by the seat of one's pants would be if flying.

 

[Stosh]

Right thumb, or left thumb

use my right thumb to replicate a Darwin, left thumb for a Provari

 

[Rocketman]

use my right thumb to replicate a Darwin, left thumb for a Provari

So your right thumb is your RMS thumb?

 

[Stosh]

I suppose one might call this method of vaping - vaping by the seat of your pants. Nothing at all wrong with it. But at three amps, best to have a high quality button to pulse with and batteries that can withstand some high momentary drains. Suspect there are some added risks there just like flying by the seat of one's pants would be if flying.

A good 5amp switch on my PT, using a 5v - 3amp plasma screen supply, no voltage sagging.
Was done as Proof of concept, not my everyday vape , no risks.

Extrapolate the voltage and amperage and ohms to normal operating range and the concept doesn't change. Using higher voltages with killer AW 10C batteries, the effective wattage for an enjoyable vape doesn't change. There's an instantaneous feedback loop, too hot - less button, ahhh good vaping.

The discussions of a couple tenths of volts on way or another seem silly sometimes, most vapers will adjust their hits for their perceived sweet spot without thinking about it.

 

[JW50]

When you hook a voltmeter, 4 dollar DMM, on the 20 volt DC range to an eGo driving an atty:
What does the meter read? I mean the display reading, not multiplied by duty cycle, not squared then square rooted ,
the actual meter reading?

Does it read 4.1 volts? 2 volts? 3.4 volts?
( I know what mine reads)
Just what math do you need to do to the cheap DMM reading to mathematically correct it?
Here's a shot of an ego, a carto, and super cheap DMM.
Could someone come up with an equation to mathematically correct this display reading (ignoring meter calibration inaccuracy) so I could calculate average power being delivered to the carto (3 ohm).
This meter obviously does not indicate RMS voltage. So it must be wrong. But, it is sort of square

BTW, what is the 'Crest Factor' of the eGo output into a resistive load?

Without knowing duty rate, don't think it is doable. However, if your cheapo meter is not too cheapo you might be able to do it with two readings from your meter. One from the DC scale and one from the AC scale. If PWM is present, you should get a non-zero reading from the AC scale. If a pure DC signal, AC reading should be zero. (And no correction needed - although you are still going to have to square something.) On the cheapo of cheapo's the AC read will be double the DC read. If this is case, no help in matter of average power. But if AC read is not double the DC read, conversion of the two reads to a duty rate is possible. With duty rate and the DC read translation to average power is possible (Although one still may have to square and square root here and there.)

Think what your display is showing is average voltage. Don't think that is necessarily "wrong". It just isn't RMS voltage. (Exception - if duty rate is 1, RMS voltage is same as DC voltage.)

 

[Rocketman]

So if I squared the 3.29 volt DC reading and then took the square root of that, then that would be closer to RMS?

Would a series capacitor inside the meter on the AC voltage scale explain the 6.5 volt reading I got?
Or is the eGo putting out more than 6.5 volts peak to peak?

You probably can't get a cheaper meter than this one. Knowing the cell voltage inside the eGo, the duty cycle under 3 ohm load, the sample rate and input impedance of the meter, there should be a mathematical way to calculate RMS volts from this DCV display.

Equation please>

 

[Rocketman]

I got it

THis cheap meter reads 8.4 volts on the 200 volt ACV scale when connected to an 18650 battery, and 4.21 volts on the 20 volt DCV scale. Isn't DCV=to RMS volts for a battery?

My equation is ACV/2, but only for my meter.

 

[JW50]

I got it

THis cheap meter reads 8.4 volts on the 200 volt ACV scale when connected to an 18650 battery, and 4.21 volts on the 20 volt DCV scale. Isn't DCV=to RMS volts for a battery?

My equation is ACV/2, but only for my meter.

It would seem you have cheopo of cheapo's type meter. You are simply getting double on the AC scale. But you might get a reasonably close approximation by doing your meter reads fresh off a full re-charge. With an eGo PWM type batt, the peak voltage should be close to 4.2 (recall that is what charger shows as output voltage). Make a DC read of the eGo unloaded. I did this and got 3.42 on one (with some age to it). Based on this read one can estimate the duty rate as 3.42/4.2 or 0.814. Therefore your read of 3.29, at 3 ohms, would suggest an average power of (3.29)^2/(3 ohms)/0.814 or 4.4 watts. If you where to be using that 18650 batt (no PWM), that loaded voltage that you saw as 3.29 on an eGo would likely be something like 3.7 or so. In this case the watts would be (3.7)^2/(3 ohms) or about 4.5 - 4.6 watts.

PS - Yes, if no PWM, DC volts equals RMS volts. In the eGo approximation shown above, RMS volts is about 3.29/(0.815)^.5 (3.29 divided by square root of 0.815) or about 3.64 volts. Or, approximately 1.11 times DC read.

PSS - On cheapo of cheapo, my understanding is that meter displays double the average voltage of half a cycle. On less than cheapo of cheapos (but still cheap), display is average of rectified voltage in excess of average voltage for a full cycle. For pure DC, average is the DC volts and nothing, rectified or not, is in excess of the average.

 

[Rocketman]

4.4 watts on an eGo, and 4.56 watts on an 18650, both with a 3 ohm carto.
Sure seems to be a big difference in vapor for such a little difference in watts.