L
J
John Fields
Guest
On Tue, 09 Oct 2007 10:12:20 -0700, "louis.brinson@gvltec.edu"
<louis.brinson@gvltec.edu> wrote:
It depends on the capacity of your battery and the efficiency of
your conversion.
--
JF
<louis.brinson@gvltec.edu> wrote:
---
It depends on the capacity of your battery and the efficiency of
your conversion.
--
JF
D
David Bonnell
Guest
On Oct 9, 3:12 pm, "louis.brin...@gvltec.edu"
<louis.brin...@gvltec.edu> wrote:
Battery technology/discharge rate comes into play as well. If the
drain (2A) exceeds the recommended values you won't have a very happy
battery.
Linear regulators, although cheap, are inefficient when dropping
significant voltage. Using your 10W device in this manner will result
in ~12W of wasted power. You'd be much better off running a battery
pack with 50% less capacity (or running two of the devices in series).
You *could* use a switching regulator, which would be more efficient
but significantly more expensive than a linear equivalent. Might make
sense if you have a really large, high-capacity battery.
<louis.brin...@gvltec.edu> wrote:
Need to know the capacity of your battery to estimate runtime.
Battery technology/discharge rate comes into play as well. If the
drain (2A) exceeds the recommended values you won't have a very happy
battery.
Linear regulators, although cheap, are inefficient when dropping
significant voltage. Using your 10W device in this manner will result
in ~12W of wasted power. You'd be much better off running a battery
pack with 50% less capacity (or running two of the devices in series).
You *could* use a switching regulator, which would be more efficient
but significantly more expensive than a linear equivalent. Might make
sense if you have a really large, high-capacity battery.
J
Jamie
Guest
louis.brinson@gvltec.edu wrote:
yard numbers.
Get the know the AH, (Amp hour) of the battery.
basically the constant amount of energy in one
hour..
since you're requiring 2 amps, I guess math
wise, that would give you more than an hour of
running time. Also, take into account of the
lowered voltage your circuit is asking for which
gives you longer time to work with.
Don't ask me the % of charge decade expected when the
rated time&load on the battery has been reached. I've seen
this figure vary all over the place depending on who
makes the cells.
--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
inefficient regulators etc..., I'll just sling some barn
yard numbers.
Get the know the AH, (Amp hour) of the battery.
basically the constant amount of energy in one
hour..
since you're requiring 2 amps, I guess math
wise, that would give you more than an hour of
running time. Also, take into account of the
lowered voltage your circuit is asking for which
gives you longer time to work with.
Don't ask me the % of charge decade expected when the
rated time&load on the battery has been reached. I've seen
this figure vary all over the place depending on who
makes the cells.
--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
R
redbelly
Guest
On Oct 9, 6:38 pm, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
Where are you getting that from?
Mark
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
Huh? You're implying the OP's battery has more than 2 AH of charge.
Where are you getting that from?
Mark
J
Jamie
Guest
redbelly wrote:
--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
E
Eeyore
Guest
"louis.brinson@gvltec.edu" wrote:
begin any calculation.
Graham
Without knowing the capacity of the battery in Ah it's not possible to even
begin any calculation.
Graham
E
Eeyore
Guest
Jamie wrote:
ass as usual.
Graham
You're an utter MORON ! You're just pulling numbers out of your ignorant stupid
ass as usual.
Graham
E
Eeyore
Guest
Jamie wrote:
Graham
Do please show your calculations for that.
Graham
J
John Fields
Guest
On Wed, 10 Oct 2007 03:31:48 +0100, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:
You belong to him, do you?
--
JF
<rabbitsfriendsandrelations@hotmail.com> wrote:
---
You belong to him, do you?
--
JF
L
louis.brinson@gvltec.edu
Guest
On Oct 9, 1:48 pm, John Fields <jfie...@austininstruments.com> wrote:
calculation is about 10 - 12 hrs. I'm going by an old formula. What
do you think?
It is a 20 Ahr battery. with the 78% efficiency of the regulator my
calculation is about 10 - 12 hrs. I'm going by an old formula. What
do you think?
D
default
Guest
On Wed, 10 Oct 2007 05:01:00 -0700, "louis.brinson@gvltec.edu"
<louis.brinson@gvltec.edu> wrote:
efficiency.
A simple linear will just drop the XS voltage (like you were standing
there with a rheostat continually adjusting the voltage) in your case,
six volts at whatever current you use would be turned to heat so that
would be less than 50% efficient. The regulator would use a
relatively insignificant amount to perform that function.
A switching regulator might be able to deliver 90%+ efficiency. The
regulator pulls current from the battery in 11 volt pulses at some
relatively high current then uses an inductor and capacitor to smooth
out the voltage to the load.
With a switcher, you generate a certain amount of noise, so may
require RFI filtering. IC Switcher specifications generally have a
graph that shows efficiency as a function of input voltage and supply
current - you may only get the 97% they claim at one current level and
90% when using less current. National has a line of "simple
switchers" that are old technology but relatively efficient and easy
to apply without a lot of critical parts. (things like inductor and
cap quality can affect overall efficiency especially at higher switch
frequencies)
It isn't all that hard to design a switching regulator from the ground
up - particularly when you only want to drop voltage.
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<louis.brinson@gvltec.edu> wrote:
Switching regulators are specified (more like hyped) in percent
efficiency.
A simple linear will just drop the XS voltage (like you were standing
there with a rheostat continually adjusting the voltage) in your case,
six volts at whatever current you use would be turned to heat so that
would be less than 50% efficient. The regulator would use a
relatively insignificant amount to perform that function.
A switching regulator might be able to deliver 90%+ efficiency. The
regulator pulls current from the battery in 11 volt pulses at some
relatively high current then uses an inductor and capacitor to smooth
out the voltage to the load.
With a switcher, you generate a certain amount of noise, so may
require RFI filtering. IC Switcher specifications generally have a
graph that shows efficiency as a function of input voltage and supply
current - you may only get the 97% they claim at one current level and
90% when using less current. National has a line of "simple
switchers" that are old technology but relatively efficient and easy
to apply without a lot of critical parts. (things like inductor and
cap quality can affect overall efficiency especially at higher switch
frequencies)
It isn't all that hard to design a switching regulator from the ground
up - particularly when you only want to drop voltage.
--
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D
David Bonnell
Guest
Assuming a "perfect" linear regulator, a constant 11.1V, and 100%
battery discharge, you'll get 10 hours from your 20 Ah battery. Your
device consumes 10 W, and the regulator will dissipate 12.2 W as
heat. Your linear regulator works out to be about 45% efficient.
An ideal 90% efficient switching regulator will only dissipate 1.1 W,
and the effective current draw on your battery reduces to 1 A instead
of 2 A. The result is (idealized) 20 hours of runtime. This would be
equivalent to using a 5 V, 40 Ah battery pack. At 66% efficiency, the
runtime decreases to about 15 hours.
Real-life runtime will not match these idealized conditions. Battery
chemistry is likely the most significant factor. What type of battery
are you running? Lithium-ion, NiCd, Lead acid, etc.?
These are somewhat helpful with respect to regulators (and
efficiency):
http://www.maxim-ic.com/appnotes.cfm/appnote_number/751/
http://www.rason.org/Projects/swregdes/swregdes.htm
L
louis.brinson@gvltec.edu
Guest
On Oct 9, 2:07 pm, David Bonnell <dbonn...@gmail.com> wrote:
20 amp hour 11.1 volt 78 % efficiency on the conversion.
J
John Fields
Guest
On Wed, 10 Oct 2007 05:01:00 -0700, "louis.brinson@gvltec.edu"
<louis.brinson@gvltec.edu> wrote:
Your load is dissipating:
P2 = I2 E2 = 2A * 5V = 10 watts,
so if your regulator is 78% efficient, then the battery is
supplying:
P2
P1 = ------ = 12.82 watts
0.78
into the input of the regulator.
With a constant 11.1V out of the battery, then, it will have to
supply:
P1 12.8W
I1 = ---- = ------- = 1.153 amperes
E1 11.1V
into the input of the regulator.
Now, since you have a 20AH battery, its time to discharge (to
terminal voltage) will be:
20AH
t = -------- ~ 17.35 hours.
1.153A
However, there's a catch...
The capacity of the battery will depend on the rate of discharge,
and if that rate is exceeded capacity will decrease.
For example, some lead-acid batteries are rated at C/10, so if your
battery was one of those and it was rated for 20AH at C/10, then
you'd get 20AH out of it if your current draw was 2 amps for 10
hours. Or more, if the drain was less.
Since your battery voltage is 11.1V it's probably not lead-acid, but
no matter what the chemistry, you have to watch the rate of drain in
order to get C.
--
JF
<louis.brinson@gvltec.edu> wrote:
---
Your load is dissipating:
P2 = I2 E2 = 2A * 5V = 10 watts,
so if your regulator is 78% efficient, then the battery is
supplying:
P2
P1 = ------ = 12.82 watts
0.78
into the input of the regulator.
With a constant 11.1V out of the battery, then, it will have to
supply:
P1 12.8W
I1 = ---- = ------- = 1.153 amperes
E1 11.1V
into the input of the regulator.
Now, since you have a 20AH battery, its time to discharge (to
terminal voltage) will be:
20AH
t = -------- ~ 17.35 hours.
1.153A
However, there's a catch...
The capacity of the battery will depend on the rate of discharge,
and if that rate is exceeded capacity will decrease.
For example, some lead-acid batteries are rated at C/10, so if your
battery was one of those and it was rated for 20AH at C/10, then
you'd get 20AH out of it if your current draw was 2 amps for 10
hours. Or more, if the drain was less.
Since your battery voltage is 11.1V it's probably not lead-acid, but
no matter what the chemistry, you have to watch the rate of drain in
order to get C.
--
JF
L
louis.brinson@gvltec.edu
Guest
On Oct 10, 1:42 pm, John Fields <jfie...@austininstruments.com> wrote:
I need to know, Thanks a million.
Dang I wish I went to the college you did. You've told me everything
I need to know, Thanks a million.
D
default
Guest
On Wed, 10 Oct 2007 08:51:03 -0700, "louis.brinson@gvltec.edu"
<louis.brinson@gvltec.edu> wrote:
place? Do the math.
In the real world we have to deal with battery chemistry, temperature,
regulator efficiency (against a whole slew of parameters) and: oh
yes, the LOAD - where it drops out, how it drops out, etc. etc.
GVTEK.EDU?????????
Learn to think. Knowledge is good, but understanding is better,
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<louis.brinson@gvltec.edu> wrote:
If you believe that . . . what's the purpose in posting in the first
place? Do the math.
In the real world we have to deal with battery chemistry, temperature,
regulator efficiency (against a whole slew of parameters) and: oh
yes, the LOAD - where it drops out, how it drops out, etc. etc.
GVTEK.EDU?????????
Learn to think. Knowledge is good, but understanding is better,
--
----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
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D
default
Guest
On Wed, 10 Oct 2007 03:31:48 +0100, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:
snip
question is either posted with very limited knowledge of electronics,
or he's just a chameleon troll.
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<rabbitsfriendsandrelations@hotmail.com> wrote:
snip
Easy there. This guy has all the hallmarks of a Google Grooper and his
question is either posted with very limited knowledge of electronics,
or he's just a chameleon troll.
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R
redbelly
Guest
On Oct 9, 8:10 pm, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
that would give you more than an hour of running time" implies a 2 Amp-
hour capacity. If you're not assuming this, how did you come up with
the 1 hour run time?
At any rate, it's a moot point now that the OP has told us it's a 20 A-
hr battery.
Regards,
Mark
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
The battery is to run at 2 amps. You're statement that "math wise,
that would give you more than an hour of running time" implies a 2 Amp-
hour capacity. If you're not assuming this, how did you come up with
the 1 hour run time?
At any rate, it's a moot point now that the OP has told us it's a 20 A-
hr battery.
Regards,
Mark
J
John Fields
Guest
On Wed, 10 Oct 2007 13:13:54 -0700, "louis.brinson@gvltec.edu"
<louis.brinson@gvltec.edu> wrote:
Very much my pleasure.
--
JF
<louis.brinson@gvltec.edu> wrote:
---
Very much my pleasure.
--
JF