Ideas for battery powered heater wanted.
 Chris Smith
 Posts: 4325
 Joined: Tue Dec 04, 2001 1:01 am
 Location: Bieber Ca.
Hi there,
According to the data sheet posted by dyrkr it looks like
a 40 ohm resistor could do the job in roughly 1 hours time.
It would deliver approx 1.2 watts over 1 hour.
I'd definitely try it first though.
The average voltage would be about 6.9 volts and the
average current about 172ma (over the discharge period of
1 hour).
Let's get our battery ratings straight first though...
A battery is rated in "mAh", which stands for 'milliampere hours"
or mathematically "milliamps * hours". Note the multiplication.
This is not the same as mA/h. You use division when you want
to find out something else about the battery such as max mA within
a given time period, or how long you can discharge at a given current.
For example, say we have a 9v battery rated at 440 mAh.
First, we want to know how long the battery will last (approx) if
we discharge at a rate of 100mA. To get this, we divide 440 by 100:
440/100=4.4 . To get the units, we divide the text "mA*h" by
"mA"...which looks like this:
?=mAh/mA .
From this we can see that the mA cancel, leaving us with:
?=h/1
which is the same as
?=h
so we know the result is in hours.
This means we can write the above like this:
h=440mAh/100mA
so we get
h=4.4 hours.
Now lets say we want to know the max rate of discharge over
a 1 hour period...to get this we simply divide by 1 hour:
?=440mAh/1h
Here, the "h's" cancel and we are left with:
?=440mA
or
I=440ma (current)
This tells us that we'll get about 440ma for 1 hour, but in practice
these kind of batteries are usually not run this hard, so we must also
look to see what the 'recommended' max discharge is for a given
battery is too, which is usually much lower than this.
BTW, this rate (for 1 hour period) is called the "1C rate" where
C represents the capacity of the battery, because in this case
440mAh is the capacity of the battery, and the '440' in the current
number (440ma for 1 hour) is the same as the '440' in the capacity
number (440mAh).
To find the "2C rate" we would divide 440mA by 2.
A lot of batteries run best at 10C rate, which would be 44mA for this
battery.
The other thing to consider is how the voltage changes over time while
the battery is being discharged. This helps to estimate the energy taken
from the battery over a given time period, as well as how much might be
left.
With a fixed resistor (say 40 ohms) the initial voltage will be about 7.2
volts (might be a little higher too) and after about 1 hour the voltage will
be about 6.6 volts, while the initial current will be about 180ma and the
final current about 165ma. To find the power we would integrate, but
to keep this simple we will approximate by the averages. The average
voltage is 6.9 volts, and the average current is 0.1725 amps.
Multiplying volts times amps (for an approximation only when the quantities
dont change by much) we get 1.19 watts. Thus, over a 1 hour period
about 1.2 watts will be dissipated in the 40 ohm resistor.
In practice there are a lot of variables that we have only approximated
here. A test would have to be conducted to make sure 40 ohms works ok.
Also, using two batteries and two 40 ohm resistors (separate circuits)
with both resistors under water (as well as insulated) would heat the
water in half the time, so a second circuit could be used as compensation
in a feedback loop that calculates time remaining to reach 99 deg F
(using a microprocessor). This would mean the water could be heated
in the exact time needed even if things change in the field, such as
initial water temperature, evaporation rate, ambient temperature,
and actual resistor resistance, etc.
I have no idea yet how accurate this 'run' has to be though.
According to the data sheet posted by dyrkr it looks like
a 40 ohm resistor could do the job in roughly 1 hours time.
It would deliver approx 1.2 watts over 1 hour.
I'd definitely try it first though.
The average voltage would be about 6.9 volts and the
average current about 172ma (over the discharge period of
1 hour).
Let's get our battery ratings straight first though...
A battery is rated in "mAh", which stands for 'milliampere hours"
or mathematically "milliamps * hours". Note the multiplication.
This is not the same as mA/h. You use division when you want
to find out something else about the battery such as max mA within
a given time period, or how long you can discharge at a given current.
For example, say we have a 9v battery rated at 440 mAh.
First, we want to know how long the battery will last (approx) if
we discharge at a rate of 100mA. To get this, we divide 440 by 100:
440/100=4.4 . To get the units, we divide the text "mA*h" by
"mA"...which looks like this:
?=mAh/mA .
From this we can see that the mA cancel, leaving us with:
?=h/1
which is the same as
?=h
so we know the result is in hours.
This means we can write the above like this:
h=440mAh/100mA
so we get
h=4.4 hours.
Now lets say we want to know the max rate of discharge over
a 1 hour period...to get this we simply divide by 1 hour:
?=440mAh/1h
Here, the "h's" cancel and we are left with:
?=440mA
or
I=440ma (current)
This tells us that we'll get about 440ma for 1 hour, but in practice
these kind of batteries are usually not run this hard, so we must also
look to see what the 'recommended' max discharge is for a given
battery is too, which is usually much lower than this.
BTW, this rate (for 1 hour period) is called the "1C rate" where
C represents the capacity of the battery, because in this case
440mAh is the capacity of the battery, and the '440' in the current
number (440ma for 1 hour) is the same as the '440' in the capacity
number (440mAh).
To find the "2C rate" we would divide 440mA by 2.
A lot of batteries run best at 10C rate, which would be 44mA for this
battery.
The other thing to consider is how the voltage changes over time while
the battery is being discharged. This helps to estimate the energy taken
from the battery over a given time period, as well as how much might be
left.
With a fixed resistor (say 40 ohms) the initial voltage will be about 7.2
volts (might be a little higher too) and after about 1 hour the voltage will
be about 6.6 volts, while the initial current will be about 180ma and the
final current about 165ma. To find the power we would integrate, but
to keep this simple we will approximate by the averages. The average
voltage is 6.9 volts, and the average current is 0.1725 amps.
Multiplying volts times amps (for an approximation only when the quantities
dont change by much) we get 1.19 watts. Thus, over a 1 hour period
about 1.2 watts will be dissipated in the 40 ohm resistor.
In practice there are a lot of variables that we have only approximated
here. A test would have to be conducted to make sure 40 ohms works ok.
Also, using two batteries and two 40 ohm resistors (separate circuits)
with both resistors under water (as well as insulated) would heat the
water in half the time, so a second circuit could be used as compensation
in a feedback loop that calculates time remaining to reach 99 deg F
(using a microprocessor). This would mean the water could be heated
in the exact time needed even if things change in the field, such as
initial water temperature, evaporation rate, ambient temperature,
and actual resistor resistance, etc.
I have no idea yet how accurate this 'run' has to be though.
LEDs vs Bulbs, LEDs are winning.
This is very interesting but I'm not certain it isn't a little over my head because I get a completely different read of the Panasonic data sheet. On the graph which showed battery life against load resistance it shows voltage curves of 5.4 to 7.2 volts  I don't really understand where these are derived from  however  I read the spec sheet to say that if the volts were 5.4 with 40 Ohm load the life would be 2.8 hours and, if it were 7.2 volts, the life would be 0.4 hours. I don't really get that ! What I do get is that the spec sheet says that the battery internal resistance is 2.76 Ohms which, with our 40 Ohm load added would produce 28.48 mA which would give a volt drop across the 40 Ohm load of about 1.14 volts and across the batt. internal R of about 7.86  which would mean that the majority of the power was being dissipated across the battery internal R and only 0.032 W across the load. If the battery only lasted 0.4 hour then that would be a total of 0.013 W.hr If it lasted 2.8 hours that would be about 0.092 W.hr Neither of thos would come anywhere near the required 1.66 W.hr Perhaps I need to learn how to read spec sheets ?
E & O E ! (How many are there ?)
E & O E ! (How many are there ?)
BB
 Chris Smith
 Posts: 4325
 Joined: Tue Dec 04, 2001 1:01 am
 Location: Bieber Ca.
Hello again,
Will, let's take a look at the voltage across the 40 ohm resistor
and the internal resistance...i am not sure how you came up
with those values but here it is:
R1=2.76
R2=40
RT=R1+R2
V1=7.2
vR1=V1*R1/RT
vR2=V1*R2/RT
I=V1/RT
where
R1 is batt internal R
R2 is our 40 ohm resistor
RT is total resistance (R1+R2)
V1 is starting voltage
vR1 is voltage across R1 when V1=starting value
vR2 is voltage across R2
I is starting current
The following numerical values were obtained by direct calculation:
R1=2.76
R2=40
RT=42.76
V1=7.2
vR1=0.464733395696913
vR2=6.73526660430309
I=0.168381665107577
From this we can see that there is a lot more voltage across
the 40 ohm resistor (R2) than across the internal resistance of
the battery (R1). This makes sense too because the 40 ohm
resistor is much higher in resistance than the internal battery R.
Checking:
vR1/R1=0.168381665107577
and
vR2/R2=0.168381665107577
which means the same current level is flowing through both
resistors, which is right for a series circuit.
Ok or no?
Will, let's take a look at the voltage across the 40 ohm resistor
and the internal resistance...i am not sure how you came up
with those values but here it is:
R1=2.76
R2=40
RT=R1+R2
V1=7.2
vR1=V1*R1/RT
vR2=V1*R2/RT
I=V1/RT
where
R1 is batt internal R
R2 is our 40 ohm resistor
RT is total resistance (R1+R2)
V1 is starting voltage
vR1 is voltage across R1 when V1=starting value
vR2 is voltage across R2
I is starting current
The following numerical values were obtained by direct calculation:
R1=2.76
R2=40
RT=42.76
V1=7.2
vR1=0.464733395696913
vR2=6.73526660430309
I=0.168381665107577
From this we can see that there is a lot more voltage across
the 40 ohm resistor (R2) than across the internal resistance of
the battery (R1). This makes sense too because the 40 ohm
resistor is much higher in resistance than the internal battery R.
Checking:
vR1/R1=0.168381665107577
and
vR2/R2=0.168381665107577
which means the same current level is flowing through both
resistors, which is right for a series circuit.
Ok or no?
LEDs vs Bulbs, LEDs are winning.

 Posts: 1752
 Joined: Fri Aug 22, 2003 1:01 am
 Location: Izmir, Turkiye; from Rochester, NY
 Contact:
On the "Typical Discharge Characteristics With Constant Resistance At 20C" graph the 5.4V, 6.0V, 6.6V and 7.2V curves are for the voltages where you consider the battery "dead".
For example, a 1K Ohm load. A fresh battery starts at 9V, the voltage decreases with amount of time the load is on. If the circuit will operate correctly until the voltage drops to 5.4V, then you can expect 90 hours of operation. If for some reason the circuit needs a minimum of 7.2V, then after only 50 hours you will need a new battery.
With a 40 Ohm load, draining the battery untill the voltage drops to 5.4V gives about 2.75 hours, if the circuit needs a minimum of 7.2V the battery life is a little more than 15 minutes. (note that the curve stops at 40 Ohms, not the 36 mentioned on page 1)
A fresh battery with a 40 Ohm load delivers:
9V / 42.76 Ohms = 0.210477A
0.210477A * 40 Ohms = 8.4190V across load
8.419V * 0.210477A = 1.772W to load
When the battery has drained down to 5.4V:
5.4V / 42.76 Ohms = 0.126286A
0.126286A * 40 Ohms = 5.05145V across load
5.05145V * 0.126286A = 0.63793W to load
0.63793W / 1.772W = 0.36, 36% of fresh battery power
When the battery has drained to 7.2V:
7.2V / 42.76 Ohms= 0.16838A
0.16838A * 40 Ohms = 6.735267V
6.735267V * 0.16838A = 1.13409W
1.13409W / 1.772W = 0.64, 64% of fresh battery power
For the Panasonic battery I now think my 30 minute life EWAG was too optimistic. One battery might heat the water one time. Hard to tell from graphs on Duracell spec linked to by Chris, but it might do a little better.
6 AAs, or better yet 6 Cs, in series make more sense for this application (IMO).

For other applications that include a voltage regulator, use the "Typical Discharge Characteristics With Constant Current At 20C" graph. For 7805s estimate a little below the 7.2V curve. For LDO 5V regulators make the estimate between the 6.0V and 6.6V curves. (Use average current of the circuit after the regulator.) ((note that this graph cuts off at 100mA. They don't expect anyone to load the battery more than this. As Chris keeps saying, you can exceed the specs if you want to. True, but I'll add, "Don't complain if you get bad results, they told you so!"))

Thanks MrAl. In my other post I didn't realize such a long explanation would be needed, and made it too short.
Cheers to all,
For example, a 1K Ohm load. A fresh battery starts at 9V, the voltage decreases with amount of time the load is on. If the circuit will operate correctly until the voltage drops to 5.4V, then you can expect 90 hours of operation. If for some reason the circuit needs a minimum of 7.2V, then after only 50 hours you will need a new battery.
With a 40 Ohm load, draining the battery untill the voltage drops to 5.4V gives about 2.75 hours, if the circuit needs a minimum of 7.2V the battery life is a little more than 15 minutes. (note that the curve stops at 40 Ohms, not the 36 mentioned on page 1)
A fresh battery with a 40 Ohm load delivers:
9V / 42.76 Ohms = 0.210477A
0.210477A * 40 Ohms = 8.4190V across load
8.419V * 0.210477A = 1.772W to load
When the battery has drained down to 5.4V:
5.4V / 42.76 Ohms = 0.126286A
0.126286A * 40 Ohms = 5.05145V across load
5.05145V * 0.126286A = 0.63793W to load
0.63793W / 1.772W = 0.36, 36% of fresh battery power
When the battery has drained to 7.2V:
7.2V / 42.76 Ohms= 0.16838A
0.16838A * 40 Ohms = 6.735267V
6.735267V * 0.16838A = 1.13409W
1.13409W / 1.772W = 0.64, 64% of fresh battery power
For the Panasonic battery I now think my 30 minute life EWAG was too optimistic. One battery might heat the water one time. Hard to tell from graphs on Duracell spec linked to by Chris, but it might do a little better.
6 AAs, or better yet 6 Cs, in series make more sense for this application (IMO).

For other applications that include a voltage regulator, use the "Typical Discharge Characteristics With Constant Current At 20C" graph. For 7805s estimate a little below the 7.2V curve. For LDO 5V regulators make the estimate between the 6.0V and 6.6V curves. (Use average current of the circuit after the regulator.) ((note that this graph cuts off at 100mA. They don't expect anyone to load the battery more than this. As Chris keeps saying, you can exceed the specs if you want to. True, but I'll add, "Don't complain if you get bad results, they told you so!"))

Thanks MrAl. In my other post I didn't realize such a long explanation would be needed, and made it too short.
Cheers to all,
Dale Y
Mr Al,
Of course you are right !  I was doing the calculations on an Excel spreadsheet and I entered the internal resistance of the battery as 276 Ohms and not 2.76. Must be either old age creeping up on me or excessive Independance Day libations. As Robert Burns the famed Scots bard once said " Is it wise to remain silent and be thought a fool ? Or to speak up and dispel all doubt. "
Dyarker  Thanks for the info ! I learned something practical and useful.
Doing the calc again and using your info. I figured that, if a 40 Ohm load was connected then, after about 1.2 hours (72 minutes) the voltage would have dropped to about 6.7 and the requisite energy of 1.66 Watt.Hours would have been added to the water. Haven't done quality control checking on that calc. yet  Will I get a different answer ?
Of course you are right !  I was doing the calculations on an Excel spreadsheet and I entered the internal resistance of the battery as 276 Ohms and not 2.76. Must be either old age creeping up on me or excessive Independance Day libations. As Robert Burns the famed Scots bard once said " Is it wise to remain silent and be thought a fool ? Or to speak up and dispel all doubt. "
Dyarker  Thanks for the info ! I learned something practical and useful.
Doing the calc again and using your info. I figured that, if a 40 Ohm load was connected then, after about 1.2 hours (72 minutes) the voltage would have dropped to about 6.7 and the requisite energy of 1.66 Watt.Hours would have been added to the water. Haven't done quality control checking on that calc. yet  Will I get a different answer ?
BB
 Chris Smith
 Posts: 4325
 Joined: Tue Dec 04, 2001 1:01 am
 Location: Bieber Ca.
 Chris Smith
 Posts: 4325
 Joined: Tue Dec 04, 2001 1:01 am
 Location: Bieber Ca.
 Chris Smith
 Posts: 4325
 Joined: Tue Dec 04, 2001 1:01 am
 Location: Bieber Ca.
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