1.
Originally Posted by STEAM_P0WERED 
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.
It doubles the input
voltage, this is a regulated mod.
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2. 09-10-2014, 07:23 PM#108
Join Date
Apr 2014
Location
Califronia, USA
Posts
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Originally Posted by mitro
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.
3. 09-10-2014, 07:47 PM#109
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Feb 2014
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FL, USA
Posts
183
Originally Posted by mitro
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.
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...
4. 09-10-2014, 08:47 PM#110
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Feb 2014
Location
FL, USA
Posts
183
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.
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