Basic DC theory question: Measuring power draw over time

Reply

  #1  
Old 10-26-07, 05:24 AM
core's Avatar
Member
Thread Starter
Join Date: Jan 2007
Location: Des Moines, IA
Posts: 1,127
Received 0 Votes on 0 Posts
Basic DC theory question: Measuring power draw over time

I've been measuring the total power draw from a 12V power source (battery) over time, but have been failing to take into account the gradual voltage drop. I'd like to increase my accuracy a bit but I'm not sure how best to go about it.

Unfortunately my multimeter isn't a clamp-on, so I can really only precisely measure the current draw at the beginning of the test before hooking everything up properly. The lead on it says 10A unfused, 15 sec max every 10 mins, so I'm leary about leaving it hooked up for the entire 4+ hour test at about 4 amps.

The load is an invertor powering a small 30W device. I know that at the beginning of the test the DC current is 4.0A and the voltage at the posts under load is 12.16V. What I was doing was just assuming a constant 4.0A amps for the entire timed test, but that invertor is going to draw more and more current as the post voltage drops.

What would be the easiest way to measure this? What I'm really after here is amp hours. Perhaps I can just use the power in watts at the begginning? But what voltage to use? Do I use the voltage under load, or is that going to be slightly inaccurate on the low side? i.e. 12.16Vx4A is 48.64W, and I can reasonably assume that _that_ will stay constant regardless of the supply voltage. If that runs for 4 hours I've used 194.56 watthours, so with a nominal 12.0V that's 16.21 amp hours, correct?

Makes sense to me; it's gotta be more accurate than assuming a constant current draw. That is, assuming I haven't overlooked something basic.

And yeah, I realize an invertor probably isn't the best device to use in such a test. The only other thing I have that would be convenient is a 12V fan but to me that's a can of worms as well: As the voltage drops, it happily just doesn't spin as fast, using less and less power. Something like that (or a light bulb) seems like it would be twice as hard to accurately measure. Or would it?

Thanks,
-core
 
Sponsored Links
  #2  
Old 10-26-07, 06:47 AM
Member
Join Date: Sep 2002
Location: welland ontario
Posts: 7,109
Received 136 Votes on 122 Posts
Not sure what your ultimate goal is but to be as aaccurate as you are attempting would require battery specs. The battery manufacturer should have a some time voltage current draw graphs you could use.

Here is an exaple of some

http://www.batterystore.com/Odyssey/OdysseyTech.htm

This looks possibly useful also.

http://www.zrd.com/blog/esdbatfrm.html

Here is a google with some other info that might be useful.

http://www.google.ca/search?hl=en&q=...s+graphs&meta=
 
  #3  
Old 10-26-07, 07:05 AM
core's Avatar
Member
Thread Starter
Join Date: Jan 2007
Location: Des Moines, IA
Posts: 1,127
Received 0 Votes on 0 Posts
Hmmm maybe I made it sound too complicated. All I'm trying to do is measure approximately how much DC power I have used. Even glancing at an ammeter every 10 minutes would suffice, but like I said I can't really leave my cheap meter passing 4A for hours.

I'm not sure such specs really help me. And incidentally, some of those numbers are precisely what I'm trying to find out but that's a whole other matter -- for now I simply want to know how many amp-hours (or watt-hours) I have consumed while using my device.

I'm sure they make a Kill-A-Watt-type device for DC, but heck this _should_ be simple to calculate with just a cheap meter, no?

-core

Edit: Just FYI I have been in phone contact with the manufacturer, and they are of no help. They can't even tell me the Ah rating for the battery! But I don't want to further complicate things, I just need to calculate my approx power usage.
 
  #4  
Old 10-26-07, 08:23 AM
Member
Join Date: Sep 2000
Location: United States
Posts: 18,497
Received 0 Votes on 0 Posts
that invertor is going to draw more and more current as the post voltage drops
Exactly the opposite.

I would measure voltage at the beginning and end, and assume that it dropped linearly throughout the run. Test the voltage periodically during the run to see if that linear assumption is correct.

I would assume that current drops proportional to voltage. You can test that assumption too by measuring current at the end of the test.

Suppose the voltage drops 20% over the test. Voltage at the beginning is "V" and thus is 0.8V at the end. Current at the beginning is "I", and thus is 0.8I at the end. So power usage at the beginning is V*I, and at the end is 0.8V*0.8I. The kicker of course is that even though the voltage and current both drop linearly, the power drop is not linear.

Suppose the experiment takes time "T". If our assumption of linear voltage drop is correct (and you're going to test that assumption), then the voltage at any time "t" is V*(1-0.2t/T). And the current is I*(1-0.2t/T). And the power is V*(1-0.2t/T)*I**1-0.2t/T), or V*I*(1-0.4*t/T+0.04*t^2/T^2).

This is more simple than it appears, since V, I, and T are all constants, so it's a simple function of t and t^2. It's that t^2 component that makes the power non-linear.

So it you integrate that over t from 0 to T, you have your total power draw. You remember your calculus, right? The integral of t^2 is (t^3)/3, and the integral of t is (t^2)/2.
 
  #5  
Old 10-26-07, 08:40 AM
Integrator97's Avatar
Member
Join Date: Apr 2002
Location: NW Arkansas
Posts: 570
Received 0 Votes on 0 Posts
Don't know what your budget is, but you could buy a cheap panel ammeter, made for constant use. You can also get software for a pc, which can constantly monitor the amps. Of course, I would look for that first, as it may require a specific meter, or not at all but it's own interface.

Also, look at this info http://www.altronix.com/index.php?pid=4
 
  #6  
Old 10-26-07, 09:05 AM
core's Avatar
Member
Thread Starter
Join Date: Jan 2007
Location: Des Moines, IA
Posts: 1,127
Received 0 Votes on 0 Posts
that invertor is going to draw more and more current as the post voltage drops
Originally Posted by John Nelson View Post

Exactly the opposite.
Whoa, how can that be? Because the on the AC side, power must remain constant or the device would fail.

On the AC side of the invertor it was reading 32W according to the Kill-A-Watt. Actual end device is an LCD monitor, chose that because it seems pretty much like an on-or-off type of deal, no in between like a lamp. Even with a lamp though I would think that the invertor would draw whatever DC amps it required to power that bulb at 120VAC.

How can I get a constant 32W out with diminishing power coming in? Sounds like free power... I like that. This is, of course, assuming that the invertor isn't just dissipating extra power for no good reason.

And yes, during my last run the voltage was in fact fairly linear:

Hours/Voltage
---
0.0 12.16V
0.5 12.12V
1.0 12.09V
1.5 12.04V
2.0 12.01V
2.5 11.97V
3.0 11.92V
3.5 11.88V
4.0 11.83V
4.5 11.77V


No, I don't remember my three years of calculus... been WAY too long and back then I was interested in women too much. But I do understand nevertheless. Looks like it's time for me to set up for another run and get the ending current this time.

Thanks,
-core
 
  #7  
Old 10-26-07, 09:24 AM
Member
Join Date: Sep 2000
Location: United States
Posts: 18,497
Received 0 Votes on 0 Posts
A 30-watt device has no ability to keep the power consumption at 30 watts. It's not that smart. It merely presents a constant load, and a 30-watt device only remains a 30-watt device if the supply voltage remains the same. If the voltage drops, that 30-watt device suddenly becomes a 25-watt device.

You don't have to believe me. Use your ammeter to measure the current after the voltage drops and see if it is higher or lower than what it was at the beginning.

You can get a very close approximation of the energy usage (and it is indeed "energy" that you are interested in, not "power") without calculus by computing this as a step function. Just divide the testing interval into a series of smaller intervals, and assume the voltage and current is constant over the smaller interval. The more smaller intervals you use, the more accurate the approximation.
 
  #8  
Old 10-26-07, 09:55 AM
Member
Join Date: Sep 2002
Location: welland ontario
Posts: 7,109
Received 136 Votes on 122 Posts
An inverter has a circuity that probably maintains a constant output voltage over a range of input DC voltage then shuts itself off when the input voltage is too low.
 
  #9  
Old 10-26-07, 09:59 AM
core's Avatar
Member
Thread Starter
Join Date: Jan 2007
Location: Des Moines, IA
Posts: 1,127
Received 0 Votes on 0 Posts
Yep, the step approximation was precisely what I was planning on squeaking by with.

And LOL I do believe you... I just don't understand how it could be, since the inverter automagically keeps the receptacles at 121.5VAC regardless of the supply voltage. So my 30W device remains a 30W device, because it is always seeing that 121.5VAC. I would think the inverter would _have_ to be that smart, i.e. become an increasingly larger load as presented to the battery, to accomplish that.

And the Kill-A-Watt instantaneous reading did remain at 30-32W during the entire test if I remember correctly, not that I'm taking any stock in its accurracy with the modified sine wave. But it did show the same reading when powered off the mains.

Starting my test now just so I get my energy usage figure. (Why can't I say "power"? Ah perhaps "power-hours" would be correct, but that sounds silly. So "energy" it is?)

I'll post back when I'm done in about 5 hours. I'm kinda excited to hopefully see a lower current draw at the end..

-core
 
  #10  
Old 10-26-07, 10:07 AM
Member
Join Date: Sep 2000
Location: United States
Posts: 18,497
Received 0 Votes on 0 Posts
Thanks Joe and core. I wasn't quite sure how an inverter works, so some of my theory is probably invalid. Nevertheless, you're going to measure the current, so we'll soon find out. The current may indeed be higher at the end. Post back and let us know how it goes so we can all get smarter.

If that inverter is indeed as smart as you say, then the computation is easy. If you use 30 watts over four hours, then you have used 120 watt-hours of energy (0.12 KWH). The inverter and the voltage and the current all become irrelevant.

"Power" is the rate of energy usage. Power over time is energy. Your utility company charges you for energy, measured in KWH. Power is measured in KW. Power and energy are not the same thing.
 
  #11  
Old 10-26-07, 10:36 AM
core's Avatar
Member
Thread Starter
Join Date: Jan 2007
Location: Des Moines, IA
Posts: 1,127
Received 0 Votes on 0 Posts
Originally Posted by John Nelson View Post
If that inverter is indeed as smart as you say, then the computation is easy. If you use 30 watts over four hours, then you have used 120 watt-hours of energy (0.12 KWH). The inverter and the voltage and the current all become irrelevant.
Not _quite_ so easy. Because of the inefficiency of the inverter, and the fan on it (which is rpm-controlled, not a constant load BTW). I suppose you bring up an interesting point though, if I know the precise power I am using at the start, and the power I am using at the end is pretty much the same, then I know the total energy usage and the voltage and current in-between become irrelevant. That second part is still an 'if' at this stage though.

But the real reason I came back to post is.... RATS my darn meter is refusing to pass any current whatsoever on the high 10A setting now! I know I tested it a "wee bit" longer than the 15sec max it says, during my run yesterday. So I may have to wait on this test.... that really irks me. Only one fuse inside which is fine, and the 10A socked is supposedly unfused anyway. Rest of the meter is fine. ARGH.

Got plenty of parts around here, trying to think of a way to cobble up a quick and dirty ammeter. I'm sure the 200mA side still works, that could be key.
 
  #12  
Old 10-26-07, 11:46 AM
ray2047's Avatar
Member
Join Date: Mar 2006
Location: USA
Posts: 33,597
Received 13 Votes on 11 Posts
Been to long since physics 101 but if your building your own I do remember you need a shunt parallel to the ammeter. The shunt is of a known resistance to handle most of the current and the remaining small amount is what is across the meter. Actual amps is of course the total of what is passing across both.
 
  #13  
Old 10-26-07, 02:38 PM
Member
Join Date: Apr 2007
Location: Near Buffalo, NY
Posts: 4,239
Received 0 Votes on 0 Posts
A power inverter is a load device that draws what it draws. The input load shouldn't change regardless of its output. That's because most inverters compensate at the output by increasing voltage.

If you overload its rated output power, the device shuts down. If the input voltage drops below threshold, the device shuts down. Until either condition manifests, it can be treated as a linear draw.
 
  #14  
Old 10-26-07, 05:28 PM
core's Avatar
Member
Thread Starter
Join Date: Jan 2007
Location: Des Moines, IA
Posts: 1,127
Received 0 Votes on 0 Posts
Well I have zero data from my test but I have learned some important lessons this afternoon:

- I will not attempt to draw 4A through a 1 ohm 1/4 watt resistor
- A slinky does not make a good shunt resistor
- A multimeter with a selector dial can contain many bearings inside. These like to fly out when you disassemble the thing.

I would have had a fighting chance at all this but the smallest resistor I had was 1ohm... way too large. A length of metal with a known resistance would have been great but my meter doesn't go down to milliohm range That's where the slinky came in, it was a nice measureable 2.5 ohms with plenty of current capacity. But of course that caused a 2+ voltage drop under load so the inverter shut down.

What I ended up doing was putting a LONG string of jumper cables inline on the positive side to get some resistance. Total resistance measured 0.3 ohms. What I planned to do was use the voltage drop across this length to indicate current. I'd know the relative change even if I can't calculate the exact current.

Three hours later, that voltage drop is still exactly 0.33V; hasn't moved at all from when I started, but the battery voltage is way down just as expected.

It was fun trying to put together something that would work without the proper equipment, but looks like I'm gonna have to wait for a new meter.
 
  #15  
Old 10-26-07, 08:32 PM
ray2047's Avatar
Member
Join Date: Mar 2006
Location: USA
Posts: 33,597
Received 13 Votes on 11 Posts
Ammeters for after-market use in cars might be a way to go.
 
  #16  
Old 10-26-07, 08:46 PM
Member
Join Date: Nov 2004
Location: CA
Posts: 2,041
Received 0 Votes on 0 Posts
I think you are missing the basic part of the equation....time. Watts over a period of time...the measurement unit would be watt hours, or kilowatt hours. If you do not have a measureing instrument capable of integrating the time element, then you are just guessing.

On the other hand, you have a watt/hour meter...the kill-a-watt. Except for losses in the inverter, which you could at least guestimate, the kill-a-watt tells you what you need to know. You could run until output voltage decreased to whatever number you want to be your endpoint, and the k-a-w will tell you how much power has been consumed.
 
  #17  
Old 10-29-07, 04:15 PM
Banned. Rule And/Or Policy Violation
Join Date: Oct 2007
Posts: 176
Received 0 Votes on 0 Posts
Most Inverters have a internal voltage regulator that will maintain a constant 115-120 it may be rated for 300 watts at 115v with a 12v input but as the dc input lowers the inv. will maintain the voltage but sacrificing wattage capacity, do a kwh. measurement at begining of the hour and measure at the precise end of the hour and average the two
 
  #18  
Old 10-29-07, 05:38 PM
Member
Join Date: Jan 2004
Location: Oregon
Posts: 1,219
Received 0 Votes on 0 Posts
You need to be clear on what you want to measure: the watt hour capacity, or the amp hour capacity of the battery.

The watt hour capacity is a measure of the energy that the battery can deliver.

The amp hour capacity is a measure of the number of electrons that the battery can push through the circuit.

They are related in that the time integral of current gives you the amp hour capacity, and the time integral of current times voltage gives you the watt hour capacity.

If the voltage remains constant, then watt hour capacity is simply voltage * amp hour capacity...of course you've already learned that voltage changes.

I would suggest building a simply constant current load. This simplest such load is a three terminal regulator driving a resistor. Working at 12V and 4A, you will need a hefty regulator and resistor, you are dissipating 48W. If you are willing to run your tests at lower current levels, then your equipment becomes smaller and cheaper.

-Jon
 
Reply

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Thread Tools
Search this Thread
 
Ask a Question
Question Title:
Description:
Your question will be posted in: