Can Power Be Measured as V^2/R Using a Bench Meter?

[Phillipdanbury]

Hello all,

I'm using peltier modules (also called thermoelectric modules) for a student project. If you're not familiar, they are small semiconductor devices that when heated on one side and cooled on another, power flows due to the Seebeck effect.

I'd like to measure power coming from individual modules. Then, when I have many modules wired in series, I'd like to measure total power of the device.

In one paper regarding a similar device, the authors compute P = I^2 * R. The authors don't say how they measure current. (Paper: A 500 W low-temperature thermoelectric generator/ Design and experimental study)

In another paper, the authors say they measure "open circuit voltages" and "short-circuit" current to derive power. I understand the open circuit voltage part, but I don't know what short circuit current is. (Paper: Ring-structured thermoelectric module).

Can I measure voltage only and compute P = V^2 / R?

If so, does the internal resistance of my bench meter (HP 34401A multimeter) count as the resistance in my circuit or do I need to add another resistor in series? Why do other researchers prefer other methods? Is the "short circuit" method preferable?

If anyone has a reference about measuring power of a device without a load, I would be very grateful. I would like to present my device at conferences and so a reference would be great.

Thanks for reading.

 

[skeptic2]

If you know the resistance, yes you can measure the voltage and calculate power. Presumably the resistance used will be the resistance used in your actual circuit. You should be aware that there is an optimal value for the resistance to get maximum power out and you may have to try various values to find it.

 

[Phillipdanbury]

Thanks for your reply.

So, I should measure the resistance across my single module and use that resistance value when computing power for that single module? Will I also add the internal resistance of the meter from the spec sheet? If not, why not?

 

[skeptic2]

No, you should connect your load to your module and then measure the voltage across the load. Unless your load has a resistance approaching that of your meter (highly unlikely) your meter's resistance will have a negligible effect.

 

[NascentOxygen]

Can I measure voltage only and compute P = V^2 / R?

Providing that R is the resistance of an external rheostat that can handle the power, and R is known, then yes.

 

[Phillipdanbury]

No, you should connect your load to your module and then measure the voltage across the load. Unless your load has a resistance approaching that of your meter (highly unlikely) your meter's resistance will have a negligible effect.

What do I use as a load? A simple resistor? I think Oxygen is saying the same thing below. Any good references for this method? Even a textbook chapter?

Providing that R is the resistance of an external rheostat that can handle the power, and R is known, then yes.

 

[NascentOxygen]

A single module is probably a low power device, just a watt or two? So the cheapest way to test a module would be to buy 3 or 4 resistors, each 1 or 2 watts power rated, and arrange them series/parallel to give yourself 7 or 8 different resistance values for load testing.

What voltage and current is each module rated?

 

[NascentOxygen]

In another paper, the authors say they measure "open circuit voltages" and "short-circuit" current to derive power. I understand the open circuit voltage part, but I don't know what short circuit current is.

The performance of a voltage source can be summarized by a graph showing how output voltage varies as load current is varied. At one extreme you have a point where load resistance is very high (an open circuit), and at the other extreme you have a point where load resistance is very low, practically zero Ohms (i.e., a short-circuit). Useful operation lies in between these extremes.

 

[Phillipdanbury]

A single module is probably a low power device, just a watt or two? So the cheapest way to test a module would be to buy 3 or 4 resistors, each 1 or 2 watts power rated, and arrange them series/parallel to give yourself 7 or 8 different resistance values for load testing.

What voltage and current is each module rated?

Three modules:

1) 15.4 vdc, 30 amp max
2) 15.4 vdc, 10 amp max
3) 12 vdc, 5.8 amp max

I don't expect to reach the temperature difference where I will see these max values. One or two watts per module is probably realistic.

 

[NascentOxygen]

I picture them to be something like these: http://www.ebay.com/itm/12V-60W-TEC1-12706-Heatsink-Thermoelectric-Peltier-Cooling-Plate-Module-BE/122122181201

 

[tech99]

An ammeter has almost zero resistance, and if you connect it directly across the device, it will read short circuit current. If you replace the ammeter with a voltmeter, which has very high resistance, it will read open circuit voltage. If you multiply the two readings together you obtain a power rating for the device. Of course, this power rating is only a rough guide, but it is a simple test and one that is used quite often with generators such as solar panels.

 

[Phillipdanbury]

I picture them to be something like these: http://www.ebay.com/itm/12V-60W-TEC1-12706-Heatsink-Thermoelectric-Peltier-Cooling-Plate-Module-BE/122122181201

Yes, those are the modules I am using.

An ammeter has almost zero resistance, and if you connect it directly across the device, it will read short circuit current. If you replace the ammeter with a voltmeter, which has very high resistance, it will read open circuit voltage. If you multiply the two readings together you obtain a power rating for the device. Of course, this power rating is only a rough guide, but it is a simple test and one that is used quite often with generators such as solar panels.

Thanks for clarifying this. This is what I was doing originally before, but it didn't seem right when I thought about it.

Connecting a second meter to measure short circuit current along with open circuit voltage seems to change the reading. Considering my values are so small, maybe I'm better off calculating P = V^2 / R rather than measuring short circuit current? Would either of you mind posting ridiculously specific directions for how to set up my meter and resistors? I have a whole book of resistors. I can make any resistance in combination.

 

[tech99]

The ammeter and then the voltmeter are connected one after the other and not at the same time.

 

[Phillipdanbury]

The ammeter and then the voltmeter are connected one after the other and not at the same time.

My test rig applies heat from a hot water bath to one side and cold from a refrigerated bath to the other side. There's about 0.2 C variation from moment to moment. Because of this, by the time I disconnect vdc and connect adc, my delta T may change. Perhaps I'm better off with P = V^2/ R for this reason?

 

[Tom.G]

...by the time I disconnect vdc and connect adc, my delta T may change.

1) Connect ONE lead of the ammeter to a peltier output terminal.
2) Connect the volmeter across the peltier output.
3) Read the Voltage.
4) Connect the OTHER ammeter lead to the appropiate peltier output terminal. (Maybe us a pushbutton switch for this)
5) Read the Amperage.

No need to disconnect the voltmeter. When the ammeter is connected, it will be a short circuit load for the peltier module.

 

[skeptic2]

An ammeter has almost zero resistance, and if you connect it directly across the device, it will read short circuit current. If you replace the ammeter with a voltmeter, which has very high resistance, it will read open circuit voltage. If you multiply the two readings together you obtain a power rating for the device. Of course, this power rating is only a rough guide, but it is a simple test and one that is used quite often with generators such as solar panels.

I'm afraid this isn't correct. For instance suppose we have a 10 volt source with a 1 ohm internal resistance. If we put an ammeter directly across the source we get a current of 10 amps. Measuring open circuit voltage gives us 10 volts. Multiplying the two readings gives a power rating of 100 watts.

However the maximum power that can be obtained from this source occurs when a load of 1 ohm is connected. Then the voltage across a 1 ohm load is 5 volts and the power in the load is E^2/R or 25 watts, a significant difference from 100 watts.

 

[Phillipdanbury]

Needless to say, I am very confused.

If there is a most appropriate way to measure total power from the generators, I can try to get additional hardware to do this. Please let me know. If any of you have resources to help me read and understand your comments further, I'm happy to do some additional reading on the subject.

Thank you all for your comments so far.

 

[skeptic2]

Probably the easiest way to determine the power available from the modules (which by the way will vary depending on the resistance of your load) is to either connect your load or connect a power resistor, of a resistance and power rating similar to what you expect your load to be, across the Peltier modules and measure the voltage across it. The power will be E^2/R. Alternatively you could insert an ammeter in series with your load and find power by I^2*R.

 

[jim hardy]

Needless to say, I am very confused.

If there is a most appropriate way to measure total power from the generators,

Tom.G gave good advice.

You need to focus on what's actually going on. Skeptic gave a hint, i'll try to add some detail to his hint. Then you'll appreciate Tom.G's suggestion.

In EE there is a theorem, "Transfer of power is maximum when impedance of load equals impedance of source". (for DC that is)
There's another theorem , "Any circuit can be represented by an ideal voltage source in series with an impedance." That representation is called "Thevenin Equivalent Circuit".

To find the Thevenin equivalent for a DC circuit you gather two measurements:
1. Measure open circuit voltage, write that number down as V_thevenin
2. Measure short circuit current

Then you divide result of (1) by result of (2) and write that number down as R_thevenin

Then you draw a circuit with a battery of voltage V_thevenin in series with a resistor of value R_thevenin .

That is the "Thevenin Equivalent Circuit". It will behave like the device under test. Well, so long as things stay simple ie neither R_thevenin nor V_thevenin is dependent on the other.

Tom.G's instructions tell you how to find "Thevenin Equivalent Circuit" for your modules. You should do it once for each one.

Now to maximum power transfer, as mentioned by skeptic :

When R_load is zero, obviously R_thevenin gets all the power.

With some algebra you could solve for power to load as f(R_load , V_thevenin and R_thevenin ).
Nowadays folks use spreadsheets to demonstrate.

You'll find efficiency is only 50% at point of max power transfer so most machines we don't run that heavily loaded. Radio antennas we do, though, to optimize signal transfer.

Anyhow, finding the Thevenin equivalent for your peltiers will give you an idea of the power available to a load. Then you could check how close to an ideal Thevenin circuit is each peltier gizmo by connecting a couple different R_load's to it and plotting results. Try 1X, ½X and 2X R_thevenin.
Chances are you'll find it somewhat nonlinear .
You can fit any three data points with a quadratic curve, and a least squares quadratic fit to four or five would make a nice slide for a presentation ...
you could teach beginners the interplay between internal resistance , power transfer and efficiency .

hope this helps,
have fun

old jim

 

[tech99]

I'm afraid this isn't correct. For instance suppose we have a 10 volt source with a 1 ohm internal resistance. If we put an ammeter directly across the source we get a current of 10 amps. Measuring open circuit voltage gives us 10 volts. Multiplying the two readings gives a power rating of 100 watts.

However the maximum power that can be obtained from this source occurs when a load of 1 ohm is connected. Then the voltage across a 1 ohm load is 5 volts and the power in the load is E^2/R or 25 watts, a significant difference from 100 watts.

I agree with you, but for devices which are non linear, such as solar cells, which tend to act as a current source, the rough measurement seems to be a useful guide.

 

[Tom.G]

@Phillipdanbury
As @skeptic2 pointed out in post #16, the actual available power from the peltier module will be 1/4 of the value found from the open-circuit voltage and short-circuit current. Even that is an approximate value, not exact, because of nonlinear behavior and self heating of the module.

The suggestion of changing the load resistance and measuring the power at each value is a good one.
The power peak is somewhere near 1/2 the open-circuit voltage and 1/2 the short-circuit current.
For best accuracy and and fewer measurement errors, measure and write down both the current thru the load and the voltage across the load for different load values.
Then multiply the voltage by the current to yield Watts of power for each load. (To determine load resistance, divide the load voltage by the load current.)
The power should increase, reach a peak, then decrease for the various loads.
You may want to try more closely spaced load values around the peak to 'home in' on it.

For a more precise measurement you will need to stabilize the temperature across the module.

Please keep us up-to-date on your results, that way we can all learn!

 

[Phillipdanbury]

Hello all,

Thank you all for these immensely helpful suggestions. I appreciate the explanations and background. I will begin collecting some data (data that I will keep this time).

I'll report back, probably in spring.

Thank you all again.