pure capasitive load phase shift not equal 90 degrees...

recca02

Joined Apr 2, 2007
1,212
As I said before...if at all the phase angle is going to depend on the capacitance value, the resistance must have some non zero value.
 

Thread Starter

fancy102

Joined Dec 21, 2007
26
may i know wheather there is inductance load available in the market?as far as i know inductive load is quite hard to be found...is it possible to have 1Henry inductive load in the market?thanks
 

SgtWookie

Joined Jul 17, 2007
22,230
That's a pretty large inductor. It will have quite a bit of resistance and parasitic capacitance, too.

If you took a spool that was 1" in diameter with the end flanges 1" apart, and wound 28 gauge magnet wire on for 4,655 turns, 74.6 turns/layer, 62.3 layers (about 2,237.5 feet) you would have an inductor measuring about 1 Henry, and the outer layer of wire would be 2.67" in diameter. However, it would have 145.22 Ohms of resistance, and I don't know offhand how much capacitance - but it would be significant.

That's an air core inductor, of course. You could experiment with toroids, but a frequency range would be needed to select the proper core material, and current would need to be known as well.

Have a look at this page: Ferromagnetic Core Sizes
http://users.catchnet.com.au/~rjandusimports/tut_3.html
 

Thread Starter

fancy102

Joined Dec 21, 2007
26
sorry is me again...how about if my load is pure inductance?i use pure inductance value bt my output show unity power factor. is this happen because my inductance value was too small and the resistance inside the inductor higher than its inductance?for ur information, i use 6m Henry inductor conencted in series
thanks
 

SgtWookie

Joined Jul 17, 2007
22,230
You only get "pure inductors" in simulation.
There are real inductors.
There are pure inductors.
But there are no real pure inductors.
 

Thread Starter

fancy102

Joined Dec 21, 2007
26
tat mean no 90 degrees in reality?? and how about the inductance value?it influence the power factor as well same as the capasitance?
 

SgtWookie

Joined Jul 17, 2007
22,230
You may get close to 90°. You will not actually get to 90°

Perhaps if you took an extremely good conductor, and took it out beyond Pluto's orbit so that you were approaching absolute zero thus achieving superconductivity, and stretched it into a huge one-layer inductor so that there would be minimal parasitic capacitance, you would get very, very close to 90°.

But you still wouldn't get there.
 

recca02

Joined Apr 2, 2007
1,212
it influence the power factor as well same as the capasitance
true.
I think the simulation softwares tend to be a bit 'practical'...I'm getting negative electrical torque from my motor...and the other day my friend reported he noticed -ve rpm???

One more thing if a pure inductor has any capacitance, It will only have a phase shift of either + or - 90 deg...only the presence of resistance should give other values...which depend on the magnitude of both the resistance, the Inductance & capacitance...
 

Thread Starter

fancy102

Joined Dec 21, 2007
26
if pure capacitane then should have phase shift 90 or -90 degrees?why my result is based on the capasitance value i add?for example i get phase shift of 73 degrees with 1 micro Farad capasitance but i get 62 degrees for 2 micro Farad of capasitance..is this possible based on the theory??

and more thing to ask..is the phase shift must be less than 90 degrees no matter what laod i put?there cant be more than 90 degrees for the phase angle i measure right?thanks
 

recca02

Joined Apr 2, 2007
1,212
You misunderstood...If capacitance is more than inductance and NO RESISTANCE is present(not even in the wires connecting the load..NOT EVEN AS THE POWER SUPPLY'S INTERNAL RESISTANCE THEN AND THEN ONLY the phase shift is going to be 90 degree with the current leading the voltage by 90 degree.

If inductance is more than capacitance and NO RESISTANCE is present not even in the wires connecting the load..NOT EVEN AS THE POWER SUPPLY'S INTERNAL RESISTANCE THEN AND THEN ONLY the phase shift is going to be 90 degree with the current lagging the voltage by 90 degree.

If there is resistance anywhere even in the wires connecting the load..OR EVEN AS THE POWER SUPPLY'S INTERNAL RESISTANCE or anywhere else ONLY THEN will it depend on the values of capacitance an inductance in the circuit.
 

SgtWookie

Joined Jul 17, 2007
22,230
Your change in phase angle - if you're using a simulation - is likely because the simulation is (reasonably correctly) adding in the resulting additional inductance and resistance parasitics. In other words, the simulation is attempting to add in the real-world consequences of increasing the capacitance.

But if you want to get to that 90° phase shift, select an "ideal" inductor model, and plug in any value for inductance that you wish. You might even get there, if all of the rest of your components are also "theoretically ideal".

But in the real world, those components don't exist.

It's hard to "get ahold of" that fact.

But in electronics, EVERY component has parasitic resistance, capacitance and/or inductance.

Pick one of the three for it's major "feature" - it also has traces of the other two.
 

Thread Starter

fancy102

Joined Dec 21, 2007
26
u mean the simulation already add in the real world consequences?tat mean i should get the same result as the simulation in my real world project?? should i trust the simulation result??

because i would like to know the accuracy of the system. i cant define whether it is accurate or not..thanks
 

Papabravo

Joined Feb 24, 2006
21,225
Absoutely not. Simulation is not measurement, and it is unwise to expect close agreement between simulation and an actual circuit. Would you release a PCB to manufacturing based only on simulation? I certainly would want to build one before ordering more than a handfull of boards. We simulate to get an idea about behavior of new or unfamiliar circuits. The agreement, if any, between simulation and reality is unlikely to be more than one or two significant digits.
 

Thread Starter

fancy102

Joined Dec 21, 2007
26
tat means i just get a roughly idea about the signal i will get from the simulation??i cant compare my result with the simulation??then how can i know whether my system is accurate or not??
thanks
 

beenthere

Joined Apr 20, 2004
15,819
As Papabravo has indicated, the best you get from a simulation is that there is some potential for the circuit to behave as designed. Then you build a prototype to find out what did not get simulated properly. After making whatever changes it takes to make the circuit perform properly, you can go to production. Several iterations of prototyping may be necessary.

Look at production pcb boards in any equipment. Just about every one will indicate a revision level, indicating design or layout changes. With any complexity, nobody gets it all right the first time.
 

thingmaker3

Joined May 16, 2005
5,083
I trust simulation about as far as I can throw it. Breadboarding and/or prototyping is the only way to know for certain if a new gizmo will work - and how it will work.

For example: the simulator has no model for stray capacitance of the body in close proximity to the gizmo. The simulator has no model poorly chosen board traces. The simulator has no accurate model for hand-wound inductors.

Software won't go any farther than pencil & paper math will, even if it does get there more quickly.
 

Papabravo

Joined Feb 24, 2006
21,225
That means I just get a rough idea about the signal I will get from the simulation? Yes
I can't compare my result with the simulation? You can but expect there to be significant differences.
Then how can i know whether my system is accurate or not? You can only be certain if you build one and test it against your requirements.

Thanks
I've corrected your original text in the quote. You really should spend some time working on your communication skills. It will greatly enhance your credibility.
 
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