PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2021-04-12, 08:30:41
News: A feature is available which provides a place all members can chat, either publicly or privately.
There is also a "Shout" feature on each page. Only available to members.

Pages: 1 [2]
Author Topic: Measuring INPUT Power Accurately and with no Oscilloscope  (Read 27247 times)

Group: Administrator
Hero Member
*****

Posts: 3073
It's not as complicated as it may seem...
This weekend ended up being spent mostly with family, so I was not able to perform all the testing I wanted to.

I did however get everything set up, and to the point I was able to obtain some preliminary results.

Using a ~3kHz square wave (~50% duty) drive from a 555 circuit (more or less the original Ainslie circuit built using Groundloop's PCB), I was able to confirm what the Pin measurement was with ~15VDC supply input through the switching circuitry on a 10 Ohm inductive resistor.

The 100W load resistor got up to about 64ºC, and the measured Pin's using 4 different methods were as follows:

1) PSU meters: 0.68A x 15.2V = 10.3W
2) Scope:  MEAN[vbat(t) x 4icsr(t)] (4i to account for the 0.25 Ohm CSR) = 9.88W
3) Scope: MEAN[vbat(t)] x MEAN[4icsr(t)] = 9.91W
4) DMM: 15.18VDC x 158.2mVDC(Vcsr) x 4 = 9.6W

No external filtering was used for the DMM measurement, and the signals were fairly "clean" with little ringing. So far, so good.

Next step is to obtain some ringing in the 1MHz frequency range, and perform the same tests again. Stay tuned for more this week.

.99
   
Group: Professor
Hero Member
*****

Posts: 2839
This weekend ended up being spent mostly with family, so I was not able to perform all the testing I wanted to.

I did however get everything set up, and to the point I was able to obtain some preliminary results.

Using a ~3kHz square wave (~50% duty) drive from a 555 circuit (more or less the original Ainslie circuit built using Groundloop's PCB), I was able to confirm what the Pin measurement was with ~15VDC supply input through the switching circuitry on a 10 Ohm inductive resistor.

The 100W load resistor got up to about 64ºC, and the measured Pin's using 4 different methods were as follows:

1) PSU meters: 0.68A x 15.2V = 10.3W
2) Scope:  MEAN[vbat(t) x 4icsr(t)] (4i to account for the 0.25 Ohm CSR) = 9.88W
3) Scope: MEAN[vbat(t)] x MEAN[4icsr(t)] = 9.91W
4) DMM: 15.18VDC x 158.2mVDC(Vcsr) x 4 = 9.6W

No external filtering was used for the DMM measurement, and the signals were fairly "clean" with little ringing. So far, so good.

Next step is to obtain some ringing in the 1MHz frequency range, and perform the same tests again. Stay tuned for more this week.

.99

 Is the measuring circuit for method "4" above, using DMM's, as shown in the attached schematic from another thread?   I just want to make sure I got this method correct -- that you are proposing.

Also, do you have any results with frequency in the few MHz range?  would like to make sure the method works for this range also.

I have currently two DMM's, an auto-ranging Innova and a cheap Cen-Tech DMM...  Will these have the needed averaging functions built in?  (How can I tell?)

Thanks, .99...
   

Group: Administrator
Hero Member
*****

Posts: 3073
It's not as complicated as it may seem...
Hi Professor.

Yes, I essentially used that schematic. Doing the measurement that way should give you accurate results. I did not use the 10k RC filter shown, but using it will only help make your measurement more accurate.

I have not yet played with the "old" Ainslie circuit to achieve MHz noise. I am hoping to build a new version of the "self-oscillator" variety that we have been discussing over at OU. This will bring the measurements to the MHz range (1~2 MHz) so will work nicely as a test for this method.

Please try the method and let us know your results.

Thanks,

.99
   

Group: Administrator
Hero Member
*****

Posts: 3073
It's not as complicated as it may seem...
Here is a preview of the Burst Oscillator Circuit I developed in order to bench-test/prove the power measurement methods I have previously described and simulated (summarized on schema01 below). Also shown is an example wave form output in BURST mode. The oscillator can be operated in CONTINUOUS mode (constant oscillation), or for a more interesting and perhaps challenging measurement, BURST mode, depending on SW1 (yet to be shown, but shorts across RDUTY for continuous).

Note the included equation for obtaining the Pout measurement.  ;)

The parts should arrive Thursday. Then the build and tweaking.

Any questions or doubts etc., feel free to raise them; I'd be happy to address all.

Regards,
.99

PS. The TC4426 is a MOSFET driver chip, modified to provide BURST control as well as drive to the IRF840 switch.
   
Group: Guest
Hey Poynt,

Before you proposed that Pin = Vbat x Iavg ,
Sometimes ago we agreed that Pin = Avg[V(t) x I(t)], and since Vbat is always constant, then Avg[V(t) x I(t)] = Vbat x Iavg
So your method of measuring Pin for a cap/battery is correct. (notice that Iavg here is not Irms or Iavg practical, but rather Iavg algebraic)




http://www.allaboutcircuits.com/vol_2/chpt_1/3.html

For Pout, Irms^2 x RL(hot) seems like resistive heating of RL.  Does it still apply if there is induction heating going on in the circuit? 
   

Group: Administrator
Hero Member
*****

Posts: 3073
It's not as complicated as it may seem...
Hey Poynt,

Before you proposed that Pin = Vbat x Iavg ,
Sometimes ago we agreed that Pin = Avg[V(t) x I(t)], and since Vbat is always constant, then Avg[V(t) x I(t)] = Vbat x Iavg
So your method of measuring Pin for a cap/battery is correct. (notice that Iavg here is not Irms or Iavg practical, but rather Iavg algebraic)
Thanks Gibbs.



Quote
For Pout, Irms^2 x RL(hot) seems like resistive heating of RL.  Does it still apply if there is induction heating going on in the circuit? 
No, this would not apply to induction heating.

One day I plan on demonstrating both the Pin and Pout measurements, since it still seems very few believe they can be done this way.
   
Group: Guest


No, this would not apply to induction heating.



Ah I see.  So if the circuit contains induction heating, any idea on how we could calculate or measure Pout?  Do you think oscillators have decent induction heating? 

Your method of Pin is pretty straight forward.  Pout method is also straight forward if there is no induction heating or all induction heating take place on the coil itself.



 
   

Group: Administrator
Hero Member
*****

Posts: 3073
It's not as complicated as it may seem...
So if the circuit contains induction heating, any idea on how we could calculate or measure Pout?
No. Maybe ION has an idea how.

Quote
Do you think oscillators have decent induction heating?  
Sorry, I don't understand the question.
   
Group: Guest

Sorry, I don't understand the question.


I mean can oscillators like the Joule thief produce heat by induction.  I just want to have an idea of how to estimate Pout.





   
Group: Elite
Hero Member
******

Posts: 3541
It's turtles all the way down
I mean can oscillators like the Joule thief produce heat by induction.  I just want to have an idea of how to estimate Pout.

Nor do I understand the question which has many possibilities. Could you expand further with a proposed circuit of the induction heating you refer to.

Do you mean the oscillation is coupled into a shorted turn secondary as the induction heating element or are you referring to eddy current losses in the core?

I cannot guess so more info please?

To successfully measure power out all sources of power dissipation in the Blocking Oscillator must be carefully measured and compared to input power including EM radiated power.

If you think EM radiated power is too small to consider, then the thermal bridge method might be a good choice, but it is not the only method. In place of the bridged method, you can do sequential thermal testing against a control.

Check it out on my bench.
http://www.overunityresearch.com/index.php?topic=702.0


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Guest
Nor do I understand the question which has many possibilities. Could you expand further with a proposed circuit of the induction heating you refer to.

Do you mean the oscillation is coupled into a shorted turn secondary as the induction heating element or are you referring to eddy current losses in the core?

I cannot guess so more info please?

To successfully measure power out all sources of power dissipation in the Blocking Oscillator must be carefully measured and compared to input power including EM radiated power.

If you think EM radiated power is too small to consider, then the thermal bridge method might be a good choice, but it is not the only method. In place of the bridged method, you can do sequential thermal testing against a control.

Check it out on my bench.
http://www.overunityresearch.com/index.php?topic=702.0

Ah yes.  Shorted turn secondary, eddy current, EM radiation are all induction?  Secondary and eddy current seems like the same type.  EM radiation is more like electrostatic.  What I'm thinking is converting these induction into electrical energy via the same primary.  Preferably electrostatic mode.  I do have a view of this.

We have an electromagnet motor.  The firing coil is the primary.  The secondary are pick up coils.  However, why use secondary when we can use the same primary as pick up coil.  Electrostatic mode is just high voltage low current.  This means we convert electrical energy back at a low current.  Now we can do measurement with just the primary coil.  How many have I lost? lol

   
Group: Guest
...
EM radiation is more like electrostatic.
...

Terrible! Maxwell will turn in his grave.   :(
A radiation means a change in time and space, therefore "not static".

   
Group: Guest
Terrible! Maxwell will turn in his grave.   :(
A radiation means a change in time and space, therefore "not static".



You sure can say this is Gibb's radiation.  lol It may be the same or differ from Maxwell.  Remember we talked about capacitive coupling and the electric field?  The hot end emit energy and the cold end is not?  We know the high end is loop back to ground somewhere, remember? 

But I have a simple question.  If we pulse a coil at 10 amps for 1 second and transfer X energy to the wheel.  Now we pulse the coil at 1 amp for 1 second.  How many pulses it takes to transfer the same X energy to the wheel? 



   
Group: Guest
You sure can say this is Gibb's radiation.  lol It may be the same or differ from Maxwell.  Remember we talked about capacitive coupling and the electric field?  The hot end emit energy and the cold end is not?  We know the high end is loop back to ground somewhere, remember? 

But I have a simple question.  If we pulse a coil at 10 amps for 1 second and transfer X energy to the wheel.  Now we pulse the coil at 1 amp for 1 second.  How many pulses it takes to transfer the same X energy to the wheel? 

One again I guess that we face a problem of definition. We must be very precise about the notions we are talking about.
Gibb's radiation or not, same or different from Maxwell, we don't care. I quote a dictionary about "radiation":
  2. Physics
     a. Emission and propagation and emission of energy in the form of rays or waves.
     b. Energy radiated or transmitted as rays, waves, in the form of particles.
     c. A stream of particles or electromagnetic waves emitted by the atoms and molecules of a radioactive substance as a result of nuclear decay.


Radiation is something that moves, with a rate of change or a rate of flux, and therefore a radiation is not "static". A capacitive coupling is not static. There are displacement currents and a voltage change in time. And in case of a constant voltage, there is no displacement current, the charges form a static pattern and thus there is no radiation.

If you meant that there is no radiation of EM waves in a capacitive coupling related to AC currents, I agree. Nevertheless there is nothing static.

   
Group: Guest

But I have a simple question.  If we pulse a coil at 10 amps for 1 second and transfer X energy to the wheel.  Now we pulse the coil at 1 amp for 1 second.  How many pulses it takes to transfer the same X energy to the wheel? 



There's no taker on this one? Let me see...  We can apply conservation of momentum. The force times the time result in momentum of the wheel, so let F be the force of 10amps and f be the force of 1 amp, take out time for simplicity:

F = f1 + f2 + f3 +... + fn

Determine n, am I doing this right? 



   
Group: Elite
Hero Member
******

Posts: 3541
It's turtles all the way down
Your question is not phrased in the language of electrical science.

When you say pulse a coil at 10 amps for one second, I have have to ask what is the inductance of the coil, then what do you mean by pulsing at ten amps?

In an inductor the current rises as a function of time when a switch is closed to a voltage source.

Is 10 amps the average current? The saturated current? The endpoint current?

Is it pulsed from a voltage source or current source. They are very different.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Guest
Your question is not phrased in the language of electrical science.

When you say pulse a coil at 10 amps for one second, I have have to ask what is the inductance of the coil, then what do you mean by pulsing at ten amps?

In an inductor the current rises as a function of time when a switch is closed to a voltage source.

Is 10 amps the average current? The saturated current? The endpoint current?

Is it pulsed from a voltage source or current source. They are very different.

Hi ION,

Good questions.  I didn't include the inductance because I think it is not important.  More inductance would means more force on the coil, but since we could solve it without knowing the force value.  Yes, there is a rise time, but I assumed it was short to make less complication.  So I guess you can treat it as maximum constant current.  If a current source is the way to describes this, then it is. 

Thanks
   
Group: Guest
Hi ION,

Good questions.  I didn't include the inductance because I think it is not important.  More inductance would means more force on the coil, but since we could solve it without knowing the force value.  Yes, there is a rise time, but I assumed it was short to make less complication.  So I guess you can treat it as maximum constant current.  If a current source is the way to describes this, then it is. 

Thanks

With simplifications consisting in neglecting the inductance and the rise time, i.e. in removing everything concerning the time variation of the current, the question becomes: why do you want to pulse the current? A continuous current is enough and here is the result from the conditions that you proposed: a constant magnetic field is created, and all the energy is dissipated in the coil resistance. You have a permanent magnet, but not very efficient...    :)

   
Group: Elite
Hero Member
******

Posts: 3541
It's turtles all the way down
With simplifications consisting in neglecting the inductance and the rise time, i.e. in removing everything concerning the time variation of the current, the question becomes: why do you want to pulse the current? A continuous current is enough and here is the result from the conditions that you proposed: a constant magnetic field is created, and all the energy is dissipated in the coil resistance. You have a permanent magnet, but not very efficient...    :)

Well spoken ex. One of many reasons for pulsing a coil is to provide stimulus for some other process or function such as spinning a rotor, providing boost or buck operation in a switchmode converter or exciting NMR, or to provide coupling in a transformer etc.

Attempts at capturing BEMF will generally yield less power due to ohmic losses. If the coil wire had zero ohms, it would be at best a break even storage element.


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Guest
With simplifications consisting in neglecting the inductance and the rise time, i.e. in removing everything concerning the time variation of the current, the question becomes: why do you want to pulse the current? A continuous current is enough and here is the result from the conditions that you proposed: a constant magnetic field is created, and all the energy is dissipated in the coil resistance. You have a permanent magnet, but not very efficient...    :)



Hi Exn,

I don't know why you want to push a pulse motor with continuous current but you made a very good point, it's not very efficient. lol  However, it would be efficient if the magnet is big and the wheel move slow.  So let me ask again, even if it's under continuous mode, what is the time require to give the wheel the same energy?  i.e.  If you pulse 10 amps for 1 second, how many seconds needed when pulsing at 1 amp to get the same energy to the wheel. 
   
Group: Guest
Hi Exn,

I don't know why you want to push a pulse motor with continuous current but you made a very good point, it's not very efficient. lol  However, it would be efficient if the magnet is big and the wheel move slow.  So let me ask again, even if it's under continuous mode, what is the time require to give the wheel the same energy?  i.e.  If you pulse 10 amps for 1 second, how many seconds needed when pulsing at 1 amp to get the same energy to the wheel. 

You make unphysical hypotheses. You consider only the time periods during which the current is constant. Thus your approximation is that of a constant current.
In real life, you can avoid neither the time constants of the circuits nor the effects of di/dt.

In any case, how do you explain the alleged effect of the current onto the wheel?

   
Group: Guest
100 pulses or seconds.  Very good.  This is what I have calculated too.  Let's check the math. 

The electromagnet force is proportional to current squared, so we can take the force ratio of 10 and 1 amp.  10^2/1^2 = 100

100 = 1 + 1 + 1 +...+ 1.

Therefore, 100 pulses of 1 amp is needed to match 10 amps pulse.  We can also double check the energy dissipate in the coil. I^2Rt is the energy consume in the coil.

10^2Rt = 1^2RT
T = 100 seconds

The conclusion is we can pulse 10 amps for 1 sec or pulsing 1 amp for 1 sec 100 times to give the same amount of energy to the wheel.  Of course to extract energy from wheel follow the same logic.  You can extract 10 amps for 1 second or 1 amp for 100 seconds. 

Now consider this.  If you have a 12 volts car battery, would you pulse your wheel at high current or low current.  Would you extract energy at high current or low current.  It is just the exchange of current for time or frequency. 

   
Group: Guest
I've also thought of something else.

The formula for electromagnet force is not only proportional to coil current squared but also proportional to core permeability squared.  That means we can use the firing coil with high permeability and pick up coil with low permeability.  That way high current in pickup coil doesn't cause much Lenz while low current in firing coil put out a large force.  I think this is much more practical.

   
Pages: 1 [2]
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2021-04-12, 08:30:41