That boost is all bucked up

[AttyPops]

So... Can anyone point me a #'s that show ... battery lifetime for the eGo Twist type of systems? Specifically:

Does using lower voltages (Buck) on the twist result in better efficiency and thus longer life per charge than using higher voltages (Boost)? Or is it the other way around? I'm assuming boost has an efficiency loss, but that may be "made up for" because of the higher voltage which is more efficient by itself. IDK if that makes sense.

I know higher voltages are more efficient as a general rule, but when running through a boost circuit to get it is that negated? Someone with more knowledge of step-up transformers and such may pipe in here for example.

Let's say I have a 1.5 Ω atty at 3.4 volts (7.7 watts) and a 3 Ω atty at 4.8 volts (7.7 watts) puff times equal. Does one run out faster? (I know the voltage won't stay at 4.8 volts for long, but for purposes of discussion...it's an example)

Inquiring minds want to know.

As always, thanks in advance to the brain trust here.

[Rader2146]

I'm on my phone right now so posting the math is out of the question. I loosely calculated the input currect for a boost regulator using constants of 8 watts, and 3.4v input (minimum voltage for the twist?), with variables of 3.7 output, and then 4.8v output. I did not calculate for regulator efficiency, inductor loss, ripple current, ect. So this is only for reference and comparison, not to be taken for true and actual figures.

8w @ 4.8v output = 2.77A input current

8w @ 3.7v output = 2.35A input current

I'll post the math and reference when I get to a computer.

[TomCatt]

So if I'm reading this correctly, then more amps equals shorter run times. But efficiency will play a significant role in these calcs, won't it?

[Rader2146]

That is correct TomCatt. Battery capacity divided by current equals runtime. So given a 650mAh Twist at 2.77A would be:

650/2770=.234 hours or approx 14 minutes of "on time"

Again this is only approximate because of the factors not accounted for in the current calculations. However those factor that I didnt account for would only serve I good purpose when you need super accurate solutions or when comparing 2 different regulators. When comparing the same regulator at different outputs the effect of those factors are minimal for the sake of comparison. For example: The TI PTN04050 regulator only drops about 2% efficiency from 1.6A to 2.4A

[AttyPops]

So for us (much slower) catts ... buck is a bit more efficient than boost )even accounting for the gain on voltage due) to booster inefficiency. So turning down the voltage and using lower ohm stuffis SHOULD be better with a buck/boost type of unit.

OTOH, it would be better to use higher voltages on a dual cell (buck only) unit... no matter what regulator (although switching much better than linear)... due to voltage increase efficiency and less need to dissipate ANY extra as heat (regardless of method... some heat generated). Linear regulators, of course, being very wasteful as voltage goes down.

I assume the Twist is PWM modulated (akin to a switched regulator).

[Rader2146]

Hold on a minute guys, I may have fat fingered the math. I put the equations into Excel to make a comparison and came up with different numbers.

Give me a bit to figure this out.

[Rader2146]

Ok....

The tread title is correct, my boost was all bucked up.

I fat fingered the math somewhere in the first equation. The kicker is that it all seemed correct based on that I have always known, or thought I knew, about a boost circuit. The even bigger kicker is that once I octuple checked the math I realized that it doesn't matter, None of it, and Ive been wrong on many occasions.

Input current remains unchanged regardless of the output voltage. The only way that input current will change is by changing the power (watts), meaning change resistance or change input voltage. The reason that output voltage does not matter is the Law of Conservation. Not "Reduce, Reuse, and Recycle" but more like "Energy cannot be created, or destroyed, only transformed". This means that Input Power must equal Output Power+Efficiancy Losses. So with efficiency losses being fairly constant, the input power and input current will also be constant because output power has not changed.

That's the best explanation that I have right now. I have the spreadsheet with the correct formulas that I will post to my public Dropbox later. You can play with the values and see that as long as the desired power is not changed, the input current also is unchanged.

Sorry for the confusion, and sorry to anyone that I may have lead astray.

[AttyPops]

Thanks for that. But that means you're, in essence, saying that buck efficiency = boost efficiency. That would surprise me. Also, the buck process may use a switching regulator or PWM so (even though it's downstream) doesn't that reduce draw on the battery?

Now I'm even more confused (not an uncommon state, BTW).

[WillyB]

This is a question that can't really be answered

[Rader2146]

With a boost regulator, input current will always be higher than output current, however input current will not change regardless of the output voltage as long as power is unchanged. So it really doesn't matter if you use HR or LR coils. Battery drain is always the same as long as desired watts is unchanged.

With a buck regulator, input current is equal to output current. So using high resistance, and therefor low current, is good for battery life.

PWM is the reason that input and output current are equal. The current is the same on both sides, but the current is only allowed to flow for a portion of the time thus reducing the downstream [average] voltage.