[SOLVED] Energy consumption calculation using oscilloscope

Hi ,

I am not a person from electronics background , but aware of the basics - i want to calculate energy consumption of a component using a oscilloscope ( i have a teledyne lecroy oscilloscope ) .

I used a current probe in one channel , a puncture probe to measure the voltage in second channel , used a basic math function to calculate the power ( Ch 1 * Ch 2 ) = P3 , and used area under the curve for P3 to get the energy , but somehow the result is not matching with the theoretical calculation.

Can some one provide me the exact technique or some references.

thanking in anticipation.

If your signals are not DC, then you have to take into account the phase relationship (Power Factor) between the voltage and current. Look at this, for starters: https://en.wikipedia.org/wiki/Power_factor

The method works generally well. If current and voltage measurements are setup correctly the energy respectively power reading will work as well. You should give details about the measurement setup and show some waveforms.

If your signals are not DC, then you have to take into account the phase relationship (Power Factor) between the voltage and current.

Click to expand...

Multiplying voltage and current in oscilloscope will take care of power factor and give correct real power readings.

Hello FvM ,


Please find the attached waveform.

https://obrazki.elektroda.pl/7899913400_1395112212.jpg.


Thanks in anticipation.

yeah by just multiplying voltage and current will give you power, and integrating the power you will get energy.. But is there a Integration unit available on the Oscilloscope??

When measuring voltage, you may need a differential voltage if there is a differential driver or bridge.

Otherwise. there appears to be a DC offset affected the result to go negative. Make sure your null input or Zero Current and voltage is zero output by calibrating the current probe and use differential voltage circuit so that you are the voltage you need.

SunnySkyguy said:

When measuring voltage, you may need a differential voltage if there is a differential driver or bridge.

Otherwise. there appears to be a DC offset affected the result to go negative. Make sure your null input or Zero Current and voltage is zero output by calibrating the current probe and use differential voltage circuit so that you are the voltage you need.

Click to expand...

Hi, how could we say that without looking at the sensing circuit..

The voltage on Power goes negative after 1st pulse edge then returns to zero then positive. This is due to a DC offset.

Most Hall current probes have a degauss and a DC null balance..

Then the Voltage goes negative later in the signal which suggests a bridge driver.. ... that's how.

Please find the attached waveform.

Click to expand...

Doesn't your LeCroy oscilloscope support screenshots? Without the zero reference of each trace, the waveforms are meaningless.

I agree that a measurement offset (e.g. when using a current transformer instead of DC capable current probe) would invalidate the power calculation.

Hello FvM

O.K., that's at least readable. I guess the load is something like a solenoid valve? So where's the contradiction to calculated power consumption?

The F1 scaling factor is rather low, so you can't see details very good. And there are apparently offsets in both voltage and current measurements which distort the power result.

should i use any filter or something to get a better waveform ?

I tried calculating the energy theoretically , the value from theoretical and oscilloscope have a large error .

1) when the oscilloscope is integrating the power , will it consider the time scale in the screen or between the cursor?

Sorry guys , if my questions are very basic .

Thanks in advance.

Hi to find energy you have to integrate around a full period and you can multiply it... Still i cant say about offset by not looking at the circuit probe points may be they are real voltage and current.. The negative current may be the result of negative voltage and the both comes to the positive energy side..

Say karthikgandiban ,

Can you learn how to make a simple measurement? Try 0V in both channels then 1V in one channel then Swap channels make sure you get 0 Product at all times for the Product Function. Then 1V on both channels. Use the cursor over a 1 time interval and compute the V^2s product to ensure you get the same. Measuring power on reactive coils is a combination of DC conductive loss, and dynamic eddy current losses during transitions. The latter requires very fast sampling rates to ensure accuracy of product functions. THat means steps should not be visible in the waveform.

THen Take your fancy waveform and break it down into small intervals and ensure that what you see on the scope matches the results.

Finally make the output product scale equal at least 50% of full scale for best accuracy.

Right now the Product, F1, shows a scale of 500V^2/div with a result near zero with 8 insiginificant decimal places.

To me your voltage test point seems suspicious becuase it never goes to 0V . Is it properly located? If it is, you are looking at a passive element with very low loss , so you are looking at reactive voltage while charging and discharging current, so you need to maximize the swing on the display to achieve the best accuracy for net power loss.