STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[broli]

@broli,

That's OK as long as you really have power. Integral over time of IV or of I^2R for that matter is energy, however. That's the energy at the end of the time period you've chosen to integrate over. Would be interesting to ascertain in your sim what constitutes a time interval for a full cycle and then compare the energies (the values of the integrals) at the end of this period.

The period of 1 second in the integral I took should be exactly 700 000 periods. And since the integral integrates a mathematical equation surely this should be accurate?

But just in case below is the integral of 1 period ie over 1/700 000 second and the previous 1 second period ie 700 000 periods. Both give the same values.

[Omnibus]

@broli and @teslaalset,

This could be especially interesting to @teslaalset -- take a look at the last row (row 1003) of my last spreadsheet. The K1003 and L1003 cells give the energy at the end of the cycle. Now we can not only assess OU by the energy-time slope but can see it in terms of energy per se. Notice L1003/K1003 > 9 times. Mind you, here we have a guaranteed full cycle.

[teslaalset]

Neither am I but we have to get to the bottom of this. To tell you the truth I trust Excel better because I see what I'm doing and exactly how I'm integrating. Now, in this file I've used 800kHz, 10Ohms and 50pF to make the time intervals more exact with respect to frequency -- one thing we have to be especially careful about is to integrate over full periods. Check it out.

REDACTED: Just replaced it with the corrected one.

@Omnibus
I had a very close look at your Excel and found 3 possible corrections:
- You applied 3.14159 for pi. If you use 'pi()' instead, your accuracy will increase
- in column E you made a typo. 16.5 should be 15.6, as you used 15.6 volts
- in column I you used 100 ohms. this should be 10 ohms

Attached the corrected result. Do you agree?

[Omnibus]

The period of 1 second in the integral I took should be exactly 700 000 periods. And since the integral integrates a mathematical equation surely this should be accurate?

But just in case below is the integral of 1 period ie over 1/700 000 second and the previous 1 second period ie 700 000 periods. Both give the same values.

If you're using the same data as in the previous screen shot the last period there doesn't appear to be a full period. I'm talking about the I-t and V-t graph there.

[teslaalset]

@broli and @teslaalset,

This could be especially interesting to @teslaalset -- take a look at the last row (row 1003) of my last spreadsheet. The K1003 and L1003 cells give the energy at the end of the cycle. Now we can not only assess OU by the energy-time slope but can see it in terms of energy per se. Notice L1003/K1003 > 9 times. Mind you, here we have a guaranteed full cycle.

This will be significantly different due to the suggested corrections in my previous post.