Possible Overunity

[hartiberlin]

Hi Singer,
did you measure the  0.13 amp current at your 10 ohm
in series with the resistor or did you short out the resistor during this
with the ampmeter ?
It must be measured in series, when you take current readings.

It is probably easier for you to measure the voltage across ( in parallel) with
the resistor.
Also in parallel with the resistor must be the storage capacitor.
Please post the voltages.

Many thanks.

[singerxyz]

Again, having a little trouble understanding terminology, but what I did was connect the 10 ohm resistor to the positive out on the unit and then connected the positive terminal of the meter to the other side. the negative out of the unit I ran to the negative terminal of the meter. Is this series? I understand series/parallel when it comes to batteries, but with caps and resistors I get confused.

But after reading hartiberlin's post I think I know how to proceed now and will re-test tonight, and post with pix.

[linda933]

Again, having a little trouble understanding terminology, but what I did was connect the 10 ohm resistor to the positive out on the unit and then connected the positive terminal of the meter to the other side. the negative out of the unit I ran to the negative terminal of the meter. Is this series? I understand series/parallel when it comes to batteries, but with caps and resistors I get confused. But I'm a fast learner ;-)

Yep.  That's right for measuring current THROUGH a load resistor.  For voltage, you measure ACROSS the load.  If you know your load resistance exactly, you don't have to measure the current to figure out the power, by the way.

Power is E times I, Volts times Amps.  If you know Ohm's Law, you know that I=E/R and so, if you know E and R, you can find P in Watts by E times E/R or, more often stated "E squared over R".

For accurate calculation of power, you should measure the (disconnected) actual value of the resistor using your ohm meter, rather than just using the marked value.  All resistor values have a manufacturing "tolerance" associated, usually 5% or 10%, which means the actual resistance is allowed to be that much more or less than the marking.

Hope this helps.  By the way, the reason Stefan suggested getting several different resistor values is because every power source or electrical system has it's own "output impedance".  You will find that your circuitry puts out more power into some values of load resistor than others.  When you find the value that gives you the absolute most output power, you have identified the output impedance of your circuit!   My loose guess for your setup is that it will put out the most power into a several hundred ohm resistor.  Just a guess!

Linda

[linda933]

A good way to home in on this magic number to extract maximum power is to notice the following:

As you increase the ohmic value of the load resistor, you will most certainly notice an increase in the output voltage.  It will be zero if you use zero ohms (dead short) and will be maximum if you use an open load (no load...open circuit). 

For most (linear) circuits, and I believe yours will probably fall into this category, the ideal load for maximum power transfer will be very near to whatever load causes the voltage to fall to half its open circuit voltage.  Try to find that value by experimenting.  You will then be able to tell if you have hit the maximum power point by raising and lowering the value slightly from there and observing a power falloff in both directions.

Linda

[linda933]

And...just in case you are not totally overwhelmed with information yet...

Please note that the load "operating point" as it is called that gives the most power into the load will not necessarily be the operating point of highest efficiency!  Maximum power transfer generally occurs when load resistance is equal to the circuit's output impedance but maximum efficiency will probably be found at yet a different load!

Efficiency is the ratio of output power over input power, as you probably know.  Finding that load is a bit more complicated because you must test both the input and output powers and plot them on a graph as you change your load resistor.  Measuring input power WILL require a current measurement and a voltage measurement at the circuit's input terminals, since the circuit's input resistance can't be measured directly (it would have to be measured while it is running which won't work).  The input power will change each time you use a new load resistor. 

It is much more arduous to test for efficiency than for just the output power, as you have to measure at both ends of the circuitry and there is no simple rule for locating the max efficiency point as the "half-voltage" rule of thumb for finding the max power transfer point.

I better shut up now before you go nuts and throw the whole setup in the trash!  Have fun!

Linda