Current draw VS time and Power increase (2 questions)

[CodeMonk]

I'm sure this has been discussed before, but I just woke up and haven't had enough caffeine yet.

Question #1:
Say you have a pedal or something else that draws 40mA. You have a battery that reads 9.0 volts.
How long before that battery drops to 8.0 volts?, 7.0 volts.
And does the power drop increase exponentially? (Like is it faster starting from 7.0 volts than 8.0 volts?).

Question #2: How to increase power output of a battery or other power source?
Say you have a battery that is at 9.0 volts. How to increase that to 12.0 volts.
I'm pretty sure I have seen something about that. Dunno if it was hear or not.
I've seen circuits (thanks to RG) about going from +9 to -9 volts.


Thanks.

[Roobin]

As far as q2, have a look at http://geofex.com/circuits/+9_to_33.htm.

[R.G.]

Question #1:
Say you have a pedal or something else that draws 40mA. You have a battery that reads 9.0 volts.
How long before that battery drops to 8.0 volts?, 7.0 volts.
And does the power drop increase exponentially? (Like is it faster starting from 7.0 volts than 8.0 volts?).

It depends on the battery chemistry. Batteries are small electrochemical "engines" precharged with as much fuel as they'll ever have in them. They put out electricity by burning the fuel inside. When the fuel gets scarcer inside and harder to get the produced electricity to the terminals, both the voltage drops and the instantly-available current drops. Each battery chemistry/physical setup has a different curve of voltage-current versus time.

Batteries are rated in Ampere-hours, that being almost literally a count of how many electrons they can force out, since an ampere is just an insanely-large number of electrons moved. 9V batteries are rated in mA-hours, because they normally provide a smaller amount of current. 9V batteries used to be 160ma-hours, literally something like 16ma for ten hours. Today they're up at over 400ma-hour, or something like 40ma for ten hours.

The reason I used ten hours for the time is that the nominal rating of a battery is usually quoted at the discharge rate for ten hours. If you discharge them at higher rates, the amount of current-hours you get is somewhat less. So a 160ma-hr battery might only last 45 minutes at 160ma.

Batteries are held to be exhausted (for a 9V battery) when they get to 7.0V. Battery makers provide volt-time discharge curves if you do the google searches.

There is no good basis I know of for saying the curve is anything other than experimentally determined.

Question #2: How to increase power output of a battery or other power source?
Say you have a battery that is at 9.0 volts. How to increase that to 12.0 volts.
I'm pretty sure I have seen something about that. Dunno if it was hear or not.
I've seen circuits (thanks to RG) about going from +9 to -9 volts.

I think that's the wrong question.

A battery will put out X power if used at just the right discharge current. At higher or lower currents, you ultimately get less total energy (that is, volts times amps times time) out of the battery because the internal burning is less efficient at other discharges.

If you mean how do you get more VOLTAGE not power, you use a step-up converter circuit, either capacitive like in my article or inductive/transformer based. Notice that this means you get less usable POWER and ENERGY out of the battery because you spend some of it in making the converter run and losses in the converter.

Bottom line - what is it you're trying to do? More voltage? More current? More power?

[CodeMonk]

Question #1:
Say you have a pedal or something else that draws 40mA. You have a battery that reads 9.0 volts.
How long before that battery drops to 8.0 volts?, 7.0 volts.
And does the power drop increase exponentially? (Like is it faster starting from 7.0 volts than 8.0 volts?).

It depends on the battery chemistry. Batteries are small electrochemical "engines" precharged with as much fuel as they'll ever have in them. They put out electricity by burning the fuel inside. When the fuel gets scarcer inside and harder to get the produced electricity to the terminals, both the voltage drops and the instantly-available current drops. Each battery chemistry/physical setup has a different curve of voltage-current versus time.

Batteries are rated in Ampere-hours, that being almost literally a count of how many electrons they can force out, since an ampere is just an insanely-large number of electrons moved. 9V batteries are rated in mA-hours, because they normally provide a smaller amount of current. 9V batteries used to be 160ma-hours, literally something like 16ma for ten hours. Today they're up at over 400ma-hour, or something like 40ma for ten hours.

The reason I used ten hours for the time is that the nominal rating of a battery is usually quoted at the discharge rate for ten hours. If you discharge them at higher rates, the amount of current-hours you get is somewhat less. So a 160ma-hr battery might only last 45 minutes at 160ma.

Batteries are held to be exhausted (for a 9V battery) when they get to 7.0V. Battery makers provide volt-time discharge curves if you do the google searches.

There is no good basis I know of for saying the curve is anything other than experimentally determined.

Question #2: How to increase power output of a battery or other power source?
Say you have a battery that is at 9.0 volts. How to increase that to 12.0 volts.
I'm pretty sure I have seen something about that. Dunno if it was hear or not.
I've seen circuits (thanks to RG) about going from +9 to -9 volts.

I think that's the wrong question.

A battery will put out X power if used at just the right discharge current. At higher or lower currents, you ultimately get less total energy (that is, volts times amps times time) out of the battery because the internal burning is less efficient at other discharges.

If you mean how do you get more VOLTAGE not power, you use a step-up converter circuit, either capacitive like in my article or inductive/transformer based. Notice that this means you get less usable POWER and ENERGY out of the battery because you spend some of it in making the converter run and losses in the converter.

Bottom line - what is it you're trying to do? More voltage? More current? More power?

Yeah, you are correct, wrong way to say the question in my head.
Take a 9 volt source (9 volt battery in this case), and make it (through whatever circuitry) put out 12 volts.
Looks like the link Roobin posted to your article is what I am looking for.
While your article roughly doubles the voltage, could I build the circuit for 18v then just put a voltage regulator in there to limit it to 12 volts?

Thanks RG, you are a life saver.

[CodeMonk]

Another question, purely theoretical. All the parameters here are highly unlikely if not impossible.

Lets say you have 2 devices (pedals, min amps, radios, whatever), that can run at whatever voltage you hook it up to. Lets also say that these 2 devices are exactly alike and both draw the same amount of current, down to the millionth of a mW (yeah I know impossible, just bear with me).
You have 2 batteries, each with the same exact voltage (down to the millionth of a volt), same brand, type, etc.

Device #1 you power with the battery alone.
Device #2, you power with the battery via a charge pump set at say 17 volts.

Now lets say the device beeps or blows up or whatever when the voltage it is getting is 8 volts.

Which device will get to that point first?

Yeah I know, all these things are highly unlikely or impossible, but I was just curious.

Thanks

[Rob Strand]

What you are asking is will boosting the voltage using a booster circuit get more life out of the battery?  the answer is no.

In simple terms the battery has a certain amount of energy, stored as chemical energy.   Suppose your 9V circuit runs a current I, the power you are draining from the battery is 9*I.  When you boost the voltage to say 18V, suppose the circuit still draws I the power you are now drawing from the battery is 18*I ie. the battery is being discharged at twice the rate.  Most circuit will draw more current than I when powered from  from twice the rails.

If your circuit pooped out at say 8.5V, a point where the battery still had a reasonable amount of energy in it, then boosting the voltage by 20% or so may help you utilize the full energy from the battery.  You should keep in mind booster circuits are not 100% efficient, some are quite poor and the fact they are present means you are wasting energy!  These days there are many devices and techniques to maximize efficiency.

[R.G.]

OK, actually, that helps. I understand the question better now... I think.

Rob is correct - a battery has a limited amount of energy stored in it as chemical fuel.
The first law of thermodynamics states that matter and energy can never be created or destroyed, only converted in form. The way this is most often expressed in power design circles is "You can't win." meaning, you can't get more power out than you put in, either as energy or fuel.

The second law of thermodynamics states that the disorder of any system will increase with time at some rate. Systems inevitably get less useful. The way this is ususally stated is "You can't break even - you always lose, at least a little bit."

So the amount of energy you get out of a battery can only approach, but never equal the amount that's originally stored there when it's made. All you can do is get most of it out, then scrape the leavings out with more and more effort, like scraping the leavings of cake batter out of a bowl. You get more of the batter out by scraping harder and longer.

When a battery is made, the fuel is put in and the battery is closed. At first, the electricity comes out easily, but at some point the easy fuel is exhausted. That spent fuel gets in the way of the remaining fuel being "burned", so getting the last bits burned into electricity gets harder and harder.

This translates into the voltage coming out being lower, and the available current being less as the battery discharges. If you put a constant-current load on a battery, much like your hypothetical effects, and plot the voltage over time, it will look something like this. The battery voltage at the start is the unused battery voltage, which is usually a little higher than nominal. A new, fresh "9V" battery will be about 9.4V. This sags down to about 9.0V very quickly, then stays near 9V, declining only slightly  over time, for a long time. For a 9V battery, when the voltage finally gets down to about 7 - 7.5V, the voltage starts declining very quickly. In perhaps 10% of the whole battery life so far, the voltage drops to nearly 0V. If you let it "rest", the voltage will recover a bit. If you warm the battery a little, the voltage will recover a bit. But eventually it will simply not produce a useful amount of electrical power output.

If you look at how much electricity you got out over time, then the long initial discharge period where the battery goes from fresh to about 7V, then the vast amount of electrical power which is available in the battery is taken out. We can think about how we'd measure the amount of electricity you get. Electrical power is volts times amps, and we total that up as watt-hours by taking the volts times amps for some little time period and then adding up all the little time periods. Since we're making the current constant, the power in any time slice is just the voltage times our constant current. So we can plot just the voltage, maybe every ten seconds, and then the energy we got out over ten seconds is the voltage times our constant current, times ten seconds. The thing that changes is that the voltage declines, so we get less and less power every ten second interval.

If we follow the manufacturer's recommendation and throw the battery away when it gets to 7V, how much of the possible battery power are we wasting? We can now calculate that as the sum of the little time slices times the rapidly declining voltage as the battery drops from 7V to near zero. If you were to do all of that, you would find that the residual energy in the battery, the difference between tossing it at 7V and squeezing it dry like a sponge would be very small, probably well under 5%.The better the battery is made, the smaller the residual is.

If we imagine a magic voltage converter that can take in whatever voltage we feed it and make 9V out of it (that is, it's a perfect converter of mass/energy, according to the First Law), all that buys us is that last few percent of energy out of the battery. We got the vast majority out without needing a perfect converter.  It gets worse.

There is no perfect converter, according to the Second Law. All converters will waste a little energy, a payment for them working. The efficiency of the converter is the ratio of how much power you get out divided by how much you put in. It's always less than 1. Sometimes much less. Complex, well designed switching converters can get up to 95+%. Linear regulators are seldom any better than 50%. So in general, when you use a linear regulator, you kiss efficiency goodbye. Also there is the issue that all switching converters need some minimum voltage to work from, so at some point, the battery doesn't have enough voltage to power the converter.

If you try to squeeze a battery really dry, you can do it a couple of ways. One is to hook battery to converter and let 'er rip. That means that the converter runs all the time, and the output voltage is constant right up until we get down to the battery voltage that will no longer run the converter. We will get some fraction of the below-7V residual power out of the battery. But that power is probably less than 3-5% of the total battery power. If the converter is less than 95-97% efficient, it eats more power converting than you can possibly save by squeezing out the last little bit of battery juice.

Another way is to use the battery mostly on its own and only switch in the converter when the voltage gets low. Now you can get out a fair fraction of the small remaining power in the battery. But you had to build a switching converter, it has to be high efficiency, and it only works part time. That's a lot of expense and effort to get maybe 2-3% more out of your batteries.

Here's the recap:
1. Batteries have a fixed amount of energy available, depending on the chemistry and how well they're made.
2. The biggest amount of the battery life is already usable. Only a small fraction is tied up in the end of life residual.
3. You can get some of that residual life out, so the batteries will "last" a bit longer. That bit longer is quite small.
4. The amount of residual you can get out is limited, and that is dependent on how hard and cleverly you work at it.

In general, unless you really have some specific needs, use the batteries as is and toss (better: use rechargables) them when they're empty. Use converters to make voltages and connections that are not otherwise possible with a single battery, like inverting the voltage for a positive ground or doubling, things like that.

[demonstar]

Just thought I'd point out that one millionth of a Watt would usually be denoted as uW (mu followed by a W), one microwatt (1x10^-6). m followed by a W is usually considered to be a thousandth of a watt or 1 milliwatt (1x10^-3).

[R.G.]

I think he really meant a millionth of a milliwatt - that is, a nanowatt (nW).
As an aside, the chip layout guys I used to work with measured chip area in nano-acres (nAc).   

[demonstar]

I think he really meant a millionth of a milliwatt - that is, a nanowatt (nW).

Yeah you're right there R.G., I didn't read what he'd wrote carefully enough. I just get a little tired of seeing prefixes incorrectly used. Such as mF when they really mean uF. I guess to be totally correct I shouldn't be using 'u', I should be using the letter mu. 

[CodeMonk]

I think he really meant a millionth of a milliwatt - that is, a nanowatt (nW).
As an aside, the chip layout guys I used to work with measured chip area in nano-acres (nAc).   

Yeah, pretty much.
Since it was all theoretical, I just picked a value to use as the tolerance that would indicate that the devices, etc I was talking about were very close in value. Or something like that.