Another way of looking at the issue, of course, is to view the battery CSR as part of the power supply, just an additional internal resistance of the battery. In that case, the uncorrected reading from the voltage probe, across the battery and the resistor, is the "correct" input voltage to the rest of the circuit. But.... then you don't get to include the power dissipation in the CSR itself as output. So both your input power and your total output power go down, because your battery  CSR is effectively "wasting" power that isn't getting to the rest of the circuit.

Quote from: TinselKoala on April 08, 2013, 06:03:44 AM
No, Lawrence, your true input battery voltage is HIGHER than what your probe reads. Your probe is reading the battery and the resistor in series, not just the battery alone, and so is reading low, by the amount of the voltage drop across the resistor.

Hmm. Let me see if I can give my explanation without confusing things too badly.

Is the battery CSR to be considered part of the "power supply", or part of the circuit being powered? Since it's dissipating some power that the battery is supplying, I tend to think of it as part of the circuit. So the battery voltage that should be used for input power to the complete circuit is that which is read directly from the battery terminals without this resistor in series.

But the probe arrangement that Lawrence must use reads the battery voltage _with_ the resistor in series, and so must be reading _lower_ than the true battery voltage that we seek. Right?

What is the magnitude of this difference, and how can we correct for it? Since we know we need an answer that is Higher than what we are reading on the battery probe, we know that we have to _add_ something positive to our reading.

The difference is the voltage drop across the resistor. The true battery voltage is higher than what the probe reads, by the value of the voltage drop across the resistor, which is given directly by the "current" probe. The only problem is the negative sign of the reading from the current probe, which, as we recall, is an artefact of the way we need to position probes in this circuit.

So you take the reading from the battery probe, and ADD the _absolute value_ of the voltage drop across the resistor given by the current probe. The result gives the true battery voltage, as if the resistor wasn't there between the battery and the probe leads.

Note that this is NOT different from what .99 said. It just puts it in a different way. Subtracting a negative number is equivalent to adding its absolute value.


Of course if the resistor is considered part of the circuit, then the power dissipation in the resistor itself must be included in the circuit's total power dissipation as output.


ETA: I think your scope itself has the ability to do this "live" by selecting the Subtract function in the Math setup screens. Subtracting the voltage drop seen by the Ch2 probe from the battery-resistor voltage seen by the Ch 1 probe will yield the correct answer, because subtracting a negative is equivalent to adding a positive value. Again, this is the same thing that .99 has said and that I have explained above.

@TK or poynt99,

I am sorry to say that the explanation is adding more confusion.  The attached is an extract of the raw data from the Input CSV file.

It has three columns.  The first is the time which we can ignore in our analysis.  The second is Ch1 which represents the Input Voltage A1-A2.  The third is CH2 which represents the Input Current (A4-A3).  Please add the necessary columns and equations.

Hope that helps all to understand and do the correct analysis....

Lawrence, when you  hook a voltmeter to a battery, just by itself, you read the battery voltage, right?

Now put a light bulb in series with your battery and your voltmeter. Is the voltmeter reading higher, or lower? It is LOWER, by the value of the voltage drop across the light bulb due to its resistance. To find the true battery voltage in this case you must ADD the value of the voltage drop -- its unsigned, absolute value -- to the reading on the voltmeter.

The probe that you are using to read your INPUT BATTERY VOLTAGE is in series with a resistor. Thus it reads LOWER than the true, unresisted battery voltage. To find the true battery voltage, you must ADD the value of the voltage drop caused by the presence of the resistor. What could be plainer?

I think you can probably add your OWN column to your OWN spreadsheet to do the computation of ADDING the ABSOLUTE VALUE of the VOLTAGE DROP indicated on CH2, to the raw reading of your CH1. It does no good to have other people do work for you that you do not understand. You must strive to understand these very basic facts about voltage, current, and power measurement if you presume to claim that your measurements indicate anything special.

For every time instant:

VBatt_true = VCH1_measured + ABS(CH2_measured)

Quote from: TinselKoala on April 08, 2013, 08:14:43 AM
For every time instant:

VBatt_true = VCH1_measured + ABS(CH2_measured)

@TK,

Your explanation is correct if the circuit is pure conventional DC.  The current can only be in one direction even if there were fluctuations.  The CH2_measured  can only be in one direction. 

In the actual experiment, no matter how you measure or view it, the current (CH2_measured ) always has a positive and a negative component.  There is something not quite right in using the ABSOLUTE value.

Since both Poynt99 and PhysicsProf have used oscilloscopes to measure Input and Output Power, I would like to hear their comments before redoing all experiments.

My point of view is to use the attached spreadsheet sample but with A4-A3 connected as in the latest Board 80 arrangement in reply 485.

In any case, as far as Board 80 is concerned, the resulting COP is greater than 1 using any of the above analysis!