Accurate Measurements on pulsed system's harder than you think.

[FatBird]

Some people always try to make things too complicated.  Lol

1.  Feed the pulse train into a battery of appropriate size & voltage.
2.  Continuously monitor the battery voltage.
3.  Keep adding more Loads (maybe light bulbs) to the battery until the battery voltage stays the same.
4.  Measure the current to the Loads.
5.  Now do a simple calculation of Volts X Amps to arrive at the Wattage that your pulse train is putting out.
6.  Really, Really, Really simple folks.


                                                                                                                     .

[TinselKoala]

Some people always try to make things too complicated.  Lol

1.  Feed the pulse train into a battery of appropriate size & voltage.
2.  Continuously monitor the battery voltage.
3.  Keep adding more Loads (maybe light bulbs) to the battery until the battery voltage stays the same.
4.  Measure the current to the Loads.
5.  Now do a simple calculation of Volts X Amps to arrive at the Wattage that your pulse train is putting out.
6.  Really, Really, Really simple folks.


                                                                                                                     .

You seem to be missing the entire point of this exercise. Please watch TinMan's first and second videos and my own video on the subject.

[TinselKoala]

How dose a scope calculate power using the math trace?.
A x B ?

Yes, my scope has Math capability. I think yours does too, doesn't it?

The scope only does what it is told, though. You can multiply two traces, but for the result to be "instantaneous power" the two input traces have to be correct for the task.

OK, by Ohm's Law and going through my calculations above, if you multiply CH1 x CH2 , and CH1 is the voltage drop across the bulb, and CH2 is the current going to the bulb, you do get the Power dissipated in the bulb. This can be done in the Math trace. Then the scope's Measurements function can be used to find the Average value of this Math trace (which accounts for the Duty Cycle.)

So I set up the apparatus and the scope to do this. The results are shown below. You can see that the values are in rough agreement with what I wound up with in my manual calculations above. The differences are due to noise, spikes, offset and rounding errors, and the fact that I probably have a partial cycle on screen at the edge. But we are still very close to what we get with the manual calculation.

I think that what was causing the confusion in the first place was the use of the wrong "average" or "mean" values for the power computation. The average needs to be taken after the multiplication, rather than taking the averages first and then multiplying them.

[digitalindustry]

OK, are we all done chasing the Red Herrings now?

Here's the solution, taken from my present readings of my circuit without inductor. (The calculations take into account the temperature coefficient of resistance since the bulb's actual current and Vdrop are being used to compute its resistance.)

With capacitor connected (bulb dimmer):
The input current from the CH2 CVR is 0.082 A. The Vdrop across the bulb is 2.8 Volts. This is constant wrt time.
So the bulb's resistance is 2.8/0.082 = 34.1 ohms.  And the power being dissipated by the bulb is I²R = (0.082)² x 34.1 = 0.224 Watt.

With capacitor disconnected (bulb brighter):
The input current _during the pulse_ is 0.163 A, and the Vdrop across the bulb _during the pulse_ is 6.32 V.  So the resistance of the bulb is 6.32/0.163 = 38.8 ohms.
And the power dissipated _during the pulse_ is therefore I²R = (0.163)² x 38.8 =  1.03 Watts. BUT the _DUTY CYCLE_ of the pulse is only 31 percent, or 0.31.
Therefore the true _average power_ being dissipated by the bulb is 1.03 W x 0.31 = 0.319 Watt. The thermal lag of the filament smooths out the peak power and produces a nearly constant brightness over the entire time.

Since 0.319 Watt is more than 0.224 Watt, the bulb appears brighter when the capacitor is disconnected.

Please check my math and assumptions, and insert your own values for input current, V_drop across the bulb, and duty cycle of the pulse train.

considering you have it set up any chance of a video for the learning?

can i ask how is the duty cycle set in the setup

you are saying the input current is higher when the bulb is brighter you are saying the lower reading was/is due to the averaging measurement TM was taking and the PWM?

edit - ahh i posted this at similar time you posted above so you are saying was due to the average.

[TinselKoala]

considering you have it set up any chance of a video for the learning?

can i ask how is the duty cycle set in the setup

you are saying the input current is higher when the bulb is brighter you are saying the lower reading was/is due to the averaging measurement TM was taking and the PWM?

edit - ahh i posted this at similar time you posted above so you are saying was due to the average.

I'll probably make a video later this evening. I've been up all night and need to get some sleep now.

The duty cycle (pulse width) and frequency in my No-Bedini-Atoll circuit are set by the Function Generator driving the Gate of the mosfet. In TinMan's original Bedini SSG circuit the frequency and duty cycle are determined by the speed of the rotor and the spacing of the coil-rotor combo and the Base resistance and the characteristics of the switching transistor.

In the first few discussions the "mean" or "average" values of the current were being compared between the Cap and NoCap conditions. (My scope calls it "average" and TinMan's scope calls it "mean" but they both refer to the same thing, and "mean" is actually the more correct term.) Since the Mean current was lower when the bulb was brighter, this seemed strange, but when the Power dissipated in the bulb is actually computed, either by hand using the _correct_ input values (not the averages) or directly by the scope's Math function, the mystery is resolved and it can be seen that the lesser average (or mean) dissipated Power does produce the dimmer bulb.

Now the question becomes how does this happen electrically in the circuit? Why does connecting the capacitor produce less power dissipated by the bulb, and why does this effect depend sensitively on the resistance between the bulb and the switching element?