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Evaluating total energy usage from just DC voltage and current


stube40
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I'm trying to calculate the total energy usage of a dynamic system that is powered by a bank of lead-acid batteries.

My method is simple - log current (using current clamp) and voltage across the battery then post process the logged data -

a) For each sample point, Instantaneous Power = V*I
b) Energy used by that sample point is Power * Time Period
c) Total Energy Used in log period is the sum of all energies calculated in (b)

This works OK, but is subject to quite a bit of error if there is any offset in the current clamp at 0A. If there is any noise on the current clamp and 0A is, say, measured as 0.1A, then this creates a false Power reading from V*I and this error accumulates across the entire measurement cycle, eventually creating quite a big error.

Is there a more clever way to calculate power usage that nullifies this current-offset error? I was thinking to try and measure the drop in charge of the battery box, but for most trials we do this drop is negligible and almost unmeasureable (at least, that is, based on my understanding of LA batteries)

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Yes, nullifying the offset cures the problem. But we have many of months worth of data that needs to have this offset removed manually and that in itself is a problem for us. I haven't come up with a good way of evaluating the offset accurately with an algorithm yet. Any suggestions welcome.

In the furture, I was hoping to just avoid the offset problem completely by measuring power used in a different way - but maybe there is no better way?

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It should be fairly simple.

Measure the transducer's output when there's no current flowing and subtract that value from the current measurements taken.

How much does it drift?

If it's significant you'll have to null at regular intervals.

If drift isn't a problem then you should be able to take the offset and apply it to the data you've alreadt taken. If you load it all into a spreadsheet then it could probably be done automatically by applying a formula.

This is an analogue problem which will exist regardless of how you measure the DC current

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What kind of current levels are you talking about here?
Is there a reason it has to be a DC clamp-on vs. a shunt resistor?


We're talking about 50A @ 190V, so the shunt resistor was avoided because of power dissapation. Not only are we talking about heat generation, we're worried about losses in our system.

However, re-thinking it I think I'd rather deal with a hot resistor and losses than the problems we're currently having with the current clamp.
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We're talking about 50A @ 190V, so the shunt resistor was avoided because of power dissipation. Not only are we talking about heat generation, we're worried about losses in our system.

What do you have that draws 9500W? An electric locomotive? The heater for an olympic swimming pool?
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What do you have that draws 9500W? An electric locomotive? The heater for an olympic swimming pool?


LOL!!

You were close with the electric locomotive guess - it's a research project into locomotion based on a new type of electric motor. The power may be 9500W, but it's pulsed so overall the total energy used is fairly small in comparison. OUr system relies on large pulses hence the scary-looking numbers.
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As I said above you'll get this problem with a sense resistor because the differential op-amp used to measure the voltage across it will still have some offset, especially if you keep the voltage low to save power. You need to measure the offset and cancel it using software, which isn't hard.

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+/-10mV over waht voltage range?

If you're talking about 190V then that' not easy because it'sn early 0.005%.

If you're talking about the 50mV read from a shunt, it's 20% which is poor.

Are you talking about 10mV of offset and if so, over what voltage range?

Here's a circuit which demonstrates the effect of resistor tolerances on the common mode rejection of a differential amplifier. The differential amplifier is set to a gain of 100, which would give 5V at 50A with a 1mΩ current sense resistor. The values of R1 and R2 should be 10k and 1M respectively but they're 0.1% lower than they should be to illustrate the effect of tollerances. Even though there's no difference in the voltage on +Vin and -Vin, the output voltage is 61mV when the inputs are both at 30V.

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+/-10mV over waht voltage range?

If you're talking about 190V then that' not easy because it'sn early 0.005%.

If you're talking about the 50mV read from a shunt, it's 20% which is poor.

Are you talking about 10mV of offset and if so, over what voltage range?

Here's a circuit which demonstrates the effect of resistor tolerances on the common mode rejection of a differential amplifier. The differential amplifier is set to a gain of 100, which would give 5V at 50A with a 1mΩ current sense resistor. The values of R1 and R2 should be 10k and 1M respectively but they're 0.1% lower than they should be to illustrate the effect of tollerances. Even though there's no difference in the voltage on +Vin and -Vin, the output voltage is 61mV when the inputs are both at 30V.


Yes, we are talking 10mV over the 200V range - I admit it's a tall order..........
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An offset of 10mV or and error of 10mV?

An error of 10mV over 200V is 0.005% tolerance which is pretty unrealistic, I don't think I've seen a multimeter that good and if you can buy a meter that accurate it certainly won't be cheap.

An offset of 10mV is more realistic but you're better off doing it in software than tweaking it.

To calibrate to a tolerance of 0.005% you need a 200V reference which is also that good or a lower voltage reference and a use a 0.005% tolerance potential divider.

I could find a 0.005% tolerance resistor at my local supplier but the only value was 1k and it's expensive.

The best voltage reference I could find was 0.01% tolerance.

Because errors add up you need components with a tolerance of 0.0025% to get 0.005% which is going to prove difficult.

EDIT:
Here's a reference with a tolerance of 0.0025% which can be used for calibration.
http://www.voltagestandard.com/New_Products.html

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