Reactive power - Reactive Generator research from GotoLuc - discussion thread

[poynt99]

Luc,

This may be helpful in resolving the low impedance drive issue with your FG. It provides a few options, including Mark's buffer in an op-amp feedback loop to minimize cross-over distortion.

[gotoluc]

Thanks for the reply poynt

I'm not sure I understand why the power going into the circuit cannot be measured in the standard way?

I looked at the circuit you propose and since electronics is not my strong side it doesn't help me understand either.

So here's what I propose, instead of building a circuit to avoid a possible problem with the 50 Ohm output resistor in the SG, how about showing me a simple circuit I can build that will make a higher voltage sine wave output then the SG can in the 700 to 800KHz range. This would eliminate the SG altogether and further test the circuit capabilities.

Thanks for your help

Luc

[gyulasun]

Hi Luc,

I would suggest to replace the bulb with a real metal film resistor of some ten Ohms and measure the AC voltage across it too. You can get acceptably low inductance load resistor from several metal film types connected in parallel like your 1 Ohm was assembled but no need for using higher than 4 or 5 pieces of them as a maximum.  Say you have 5 pieces of 220 Ohm metal film or carbon type (say 1/8 or 1/4 W rated each) and you parallel them so you get 44 Ohm and can use this in formula P=V*V/44 to get output AC power (V is in rms). If you have other resistor values in the some hundred Ohm range, just use them in parallel and check their value with a digital Ohm meter.
When using a bulb as a load, one can never know the output power dissipated in it if its instanteneous current is not measured together with the voltage across it, and a further problem is the bulb repesent a varying resistor as per its brightness changes, altering its loading effect to the circuit driving it, which may cause other unwanted effects in certain circuits (like changing the loaded Q of an LC tank, for instance)

All I mean with a real resistor instead of the bulb is that output power can also be measured, beside the MATH function of the scope.

Gyula

[gotoluc]

Hi Luc,

I would suggest to replace the bulb with a real metal film resistor of some ten Ohms and measure the AC voltage across it too. You can get acceptably low inductance load resistor from several metal film types connected in parallel like your 1 Ohm was assembled but no need for using higher than 4 or 5 pieces of them as a maximum.  Say you have 5 pieces of 220 Ohm metal film or carbon type (say 1/8 or 1/4 W rated each) and you parallel them so you get 44 Ohm and can use this in formula P=V*V/44 to get output AC power (V is in rms). If you have other resistor values in the some hundred Ohm range, just use them in parallel and check their value with a digital Ohm meter.
When using a bulb as a load, one can never know the output power dissipated in it if its instanteneous current is not measured together with the voltage across it, and a further problem is the bulb repesent a varying resistor as per its brightness changes, altering its loading effect to the circuit driving it, which may cause other unwanted effects in certain circuits (like changing the loaded Q of an LC tank, for instance)

All I mean with a real resistor instead of the bulb is that output power can also be measured, beside the MATH function of the scope.

Gyula

Hi Gyula,

thanks for your post. 
I used the bulb just as a visual display to show the circuit outputs real current.  My first question was and still is,  when a circuits voltage and current is 180 degrees out of phase does this mean the input power is reflected back to the source?

Once that question is answered I would post the RMS voltage across a 10 Ohm 1% metal film resistor as load instead of the bulb.

I do know not to use bulbs to make power calculations but thanks for bringing it up.

Luc

[gyulasun]

Hi Luc,

To answer your question, yes, the output (not input) power seems to be reflected back to the source.

Here is an animation of Power Factor when the phase angle between the AC voltage and current changes between -90° and +90°.    http://demonstrations.wolfram.com/ACPowerFactorPrinciple/ 

You can read explanations in the Details section (under the 3 Snapshots) and the last sentence is: 

"If the phase angle were to be shifted to be greater than 90 degrees in either direction, the load would effectively become the source."

This means from the explanations that if the power curve is shifted in the negative direction (below the time axis as in your scope shot) then the load "delivers" power back to the generator.

Whether this is really happening in your circuit I do not know, the scope shows that to be the case. Further circuit explorations are needed.

Gyula