Power Calculation from Current Waveform

The simple answer is to measure the area under the line, however, as a battery discharges, its terminal voltage falls, which tends to complicate matters. As a first approximation, though, this figure will prove sufficient for determining the minimum battery size and multiplying by the voltage gives the approximatee power stored.

Of course, providing a capacity in excess of the theoretical requirements will avoid embarrassment.

Get a stop watch and time it.

Well, the area under the curve divided by the time will give you average current.

You could do it with a ruler. You have a few triangles and squares.

For a first approximation, you can assume that the voltage of the battery doesn't drop, but you can do voltage the same way.

Multiply them together and BINGO! - Power.

Thank You.I had imported the load current measurement data points to excel and plotted a graph of current versus time which is approximately coming in the above fashion.

Considering supply voltage to be constant 12V(to get maximum case),what I was thinking was to multiply each current data point with corresponding voltage point to get the power waveform.

Then to find the rms power take the root of mean of the squares of the points ....Should I use time in mS or the no.of data points?..I'm not sure if I'm in right direction.

As it comes from a battery, the concept of RMS is the same as "smooth DC". Don't try to make it too complicated, Eric.

I forgot to add. Everything has to be in the right units: Amps, Volts, Seconds.

You can't edit immediately after submitting even though I clicked the button.

Just a few queries before we get to a solution.

1) What are the periods for which the current rises to 40A, stays flat and fall back to 2A?

2) What is the period for the final drop off to zero current?

3) Does the current immediately rise or is there a wait duration after which this waveform repeats?

4) What is the type of load -> resistive, inductive, capacitive?

and finally 5) Is the load a controlled current sink, meaning if the voltage from the battery dropped does the current waveform remain the same or is there a proportional drop?

"What is the period for the final drop off to zero current?"

With no external intervention, infinity.

1) What are the periods for which the current rises to 40A, stays flat and fall back to 2A?

It's 5ms to 40A,10mS at 40,again 5mS to 2A

2) What is the period for the final drop off to zero current?

It's 20mS

3) Does the current immediately rise or is there a wait duration after which this waveform repeats?

There's a 150mS wait duration between pulses

4) What is the type of load -> resistive, inductive, capacitive?

It's a motor load

and finally 5) Is the load a controlled current sink, meaning if the voltage from the battery dropped does the current waveform remain the same or is there a proportional drop?

There will be proportional drop.

Thanks

From my point of view, your first step should be to draw an accurate scale drawing. Otherwise the drawing will likely mislead you. Here is your drawing followed by one made using your numbers:

Second, again in my point of view, that's NOT your waveform. to show a repetitive waveform, you MUST show at least 2 complete repetitions:

Again, without viewing at least a couple of repetitions, you don't get an accurate mental picture.

Third, I'm amazed that no one has said this sure looks like homework!

You apparently missed it, too. Look closely at op drawing the current drops to 2A not 25. time for a redraw - -

Typical ratings for battries in AH are considered to be a 2 amp drain.

In other words a 50AH battery should be expected to supply 2A for 50 hours before voltage is depleted to the minimum battery voltage allowed ( discharged level ).

Drainages are not guaranteed to be linear so do not expect the same battery to provide 4A for 25 hours.

OOPS! Coming Shortly...

Wrong!!! Where did you get that idea from?

50 hours at one amp.....that is why its called 50 amp/hours Battery....

Theoretically or 50 amps for one hour!!! Is the same....

At 2 amps, the battery should last 25 hours.....

You are too LongintheTooth to remember!!!

True, I correctd one mistake and made another of my own LOL.

If you look at some Gel, LA, and deep discharge RV battery specs though, you will see typical drain rates for AH ratings are done at 2A.

Check out the Trojan site and those are typical. I'm not sure why the industry has chosen 2A drain rates; other than to cause math errors.

In any case, you are correct that 50AH is the implied duration for 1A.

In my haste to post I got rate and capacity confused.

I stand corrected thanks.

You wrote:-

If you look at some Gel, LA, and deep discharge RV battery specs though, you will see typical drain rates for AH ratings are done at 2A.

You can actually do the drain tests at any current you wish, in fact the capacity may actually "peak" at a particular current draw.....that was not what I was being picky about!! It was your appalling math!!

Interesting manual, but limited to LA and NiCd batteries. (I did not read the entire thing...) Have you found anything equivalent for today's Li & other batteries?

For others I go to the Battery University....mostly good I find.

Thank You for the document Andy.

I hope it proves helpful.....

Thanks for looking closely enough to see my errors. Try these:

Thanks for the figures,these are the ones that I meant...

Root - sqrt()

Mean = Average

Square = ^2

That said all your I's are positive and all of you V's are positive, so V*I is power.

I is wierd, so you need the calculus definition of Average which is basically area under curve/period.

Volts, Power and Amps need everything in terms of Volts Amps and seconds.

So you have a 50A 12 volt battery. There are many types of LA battery around, you did not mention which. That makes a difference.....

If it is a normal car battery type, then these are very easily damaged if left at less than 12.6 volts for any length of time....assuming an average battery, but manufacturers data should be used. They start to sulphate.

In a car, the engine runs after starting (usually!) and recharges the battery immediately, before any real damage is done.....

The type/style of charger/charging used also plays a big role in battery life and capacity.......even some good expensive chargers can ruin a battery within a very short time when incorrectly used, wrong type of battery or charger cycle or if you simply don't know what you are doing with it!!!same/same/same!!

Best of luck!

Can you not calculate a root mean square from those crisp triangles? The RMS should be a good (useful) indication of power output over the duty cycle.

Following up on the queries for the current waveform timings and the responses got till now, here is a list of clarifications:

1) Root mean square is unnecessary as it it only DC current - there is no current going below zero in the given cycle: so average is enough - infact for DC RMS value is the same as average.

2) Drop off period cannot be infinity - that means there is only one cycle of the waveform and there is no further discharge. So, we still need drop off period and period from end of one cycle to the next to solve this.

3) Considering it is motor load, is this only the starting current or is it running current under nominal loads.

4) Is there any variations in the load over time - a typical application would be a pressure pump which draws low current when there is no pressure in the tank and builds up as the tank pressure increases. Constant load examples could be a water pump lifting water to a (open) tank at a fixed height form the ground.

5) Lastly, the battery voltage will keep dropping as it discharges and complicates the calculation, but we can get at the first approximation for this and then tackle that issue as the next step.

Awaiting your replies........

Thanks for the reply.

Regarding the load, this is the current waveform for a motor that stretches a spring(the exact waveform is a smooth curve) and releases the same .The first part is the inrush current(motor runs free) and the second part(2A) is the load current when the spring starts stretching.Full cycle is made to repeat at an interval of 150ms.

"...waveform for a motor that stretches a spring..."

If you had said that in the original post, I would have been Somewhat less likely to make the homework comment.

"Full cycle is made to repeat at an interval of 150ms."

... not according to your previous posts. It's not very clear, but I read 154 ms at the end of the main up-slope of your drawing, and you stated in post #7 that it takes 20 ms for the current to drop to zero, and that there was a 150 ms delay before the next cycle. That's a total of 324 ms before the end of the first cycle.

Now you have stated that the initial pulse is the inrush at startup. It is not at all clear whether the motor stops at the end of each pulse, or continues to run. Based on your numbers, and assuming the 2 Amps is the idling current of the motor, it takes almost twice as much energy to start the motor from scratch as it does to keep it running for 150ms idling time

Now on the other hand, you stated that the current drops to zero, indicating a full stop.

Of course we have no idea what kind of mechanism you are using to convert the rotary motion of the motor to the (presumably) linear motion of the spring (you did say 'stretch', not 'twist' or 'bend'). A cam is the first thing that comes to mind, but there are LOTS of other possibilities.

Whenever a load is suddenly removed from a motor, as may occur when your spring is released, the motor will temporarily overspin, in which case the current will not only drop rapidly to zero, but may well go negative while the motor acts as a generator.

We need much more detail before there is any hope of predicting your battery life!

My current best guess is that you want something like one or the other of these waveforms:

Thanks.Uploading a sketch with time and currents updated;sorry for the poor clarity in paint.

Is that a logarithmic scale on the vertical axis? It should be linear to correctly illustrate the power. 2A should be 1/20 of the way to 40A, and 25A should be only a little past half way...

Again, some information on the mechanism used would be helpful. Your graphs show quite a large amount of energy expended in getting the motor up to speed, and this one makes it clear that you are currently shutting the motor off between cycles.

A redesign of the mechanism could probably allow continuous run of the motor, saving much of the inrush energy.

Continuing further on the queries:

1) Does the original waveform reflect the current values at start of spring loading or the end of spring loading, or is it in the middle of the "spring cycle"?

2) Refer to reply number 14 of this thread from dkwarner. Consider the second waveform:

2a: what is the period from the top of the triangle falling to the point where the current becomes zero?

2b. what is the period between two cycles? If this cannot be defined then we may have to calculate energy required for one "spring cycle" and define battery capacity as "number of spring cycles" and not duration.

We are almost there - though my next access to the internet is couple of hours away.

ThankYou

1) Does the original waveform reflect the current values at start of spring loading or the end of spring loading, or is it in the middle of the "spring cycle"?

The waveform uploaded recently shows one spring stretch(150mS) followed by rest period

2) Refer to reply number 14 of this thread from dkwarner. Consider the second waveform:

2a: what is the period from the top of the triangle falling to the point where the current becomes zero?

20mS

2b. what is the period between two cycles? If this cannot be defined then we may have to calculate energy required for one "spring cycle" and define battery capacity as "number of spring cycles" and not duration.

It is approximately 150mS if we neglect the 0.1A dip .For the energy calculation mentioned,is the method V*I*t?

I can send this excel file if there is an email list relevant to this case. Kind of worked this off with the information collected over the past few postings.

Here is the updated calculations. Goofed on the last one with respect formula for trapezoid. Can email the file if interested.

I don't think the voltage drop is important since the OP wants to know in how time the battery will get drained and not the "power "produced [or wasted].

If we divide the under curve surface in geometric figures-as triangles and rectangles -we shall get 2340 A*ms per a cycle of 150+150=300 ms=0.3 sec that means 2.34/0.3=7.8 A*sec/1 sec=7.8 A.h/1 h. Then the 50 A.h will be drained in 50/7.8=6.41 hours.