Battery question

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It depends on the capacity of your battery and the efficiency of
your conversion.

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.

Due to other things not being equal in the equations like 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"
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Huh? You're implying the OP's battery has more than 2 AH of charge. Where are you getting that from?

Mark

I wasn't implying anything.

--
"I\'m never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

Without knowing the capacity of the battery in Ah it's not possible to even begin any calculation.

Graham

You're an utter MORON ! You're just pulling numbers out of your ignorant stupid ass as usual.

Graham

Do please show your calculations for that.

Graham

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You belong to him, do you?

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?

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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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):

20 amp hour 11.1 volt 78 % efficiency on the conversion.

--- 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

Dang I wish I went to the college you did. You've told me everything I need to know, Thanks a million.

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,

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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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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

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Very much my pleasure. :-)