Verifying the three currents add to 0

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I want to create circuit and confirm the sum of the three currents is 0
I always think of simulating the inverter phase currents
1693327853136.png

as a first step i have created the above circuit and i want to verify the currents at Neutral is 0. But it does not add to 0 as the result
1693328471898.png
 

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  • #2
Turn the inductor's polarity dots on, and you will see that L1 is backwards in the circuit compared to L2 and L3. The inductor currents then total less than 1 fA, which for me, is close enough to zero.
The three pulse sources are identical, as is the neutral voltage.
 
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  • #3
1693330911259.png

Another simulation with different duty cycle. It adds upto 0.
 
  • #4
I must take your word for it.
I'm unable to read the picture, or run the .asc
Since neutral is floating, the sum of the three currents must be zero.
Edit a plot to show the value I(L1)+I(L2)+I(L3)
 
  • #5
1693332090425.png

1693332123247.png

1693332893402.png


Sorry i am not sure the asc file is not running, through dos i renamed to txt and attached.
 

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  • #6
This is not the sort of problem you should use a simulator for. It has a simple algebraic solution. Unless the only point is to learn how to use a simulator, I guess.
 
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  • #7
DaveE said:
This is not the sort of problem you should use a simulator for. It has a simple algebraic solution. Unless the only point is to learn how to use a simulator, I guess.
Because the simulation disagreed with the obvious solution, this was a "learning to use the simulator" exercise.

PhysicsTest said:
Sorry i am not sure the asc file is not running, through dos i renamed to txt and attached.
Mosfetsim3.asc runs OK.
 
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  • #8
Also note: numerical simulators will very rarely give exact zero as an answer. That's just how these things work, numerical solution is in a way only an approximation.
 
  • #9
Borek said:
Also note: numerical simulators will very rarely give exact zero as an answer.
One least significant bit in double precision is about 9 femto units, or -320dB.

Borek said:
That's just how these things work, numerical solution is in a way only an approximation.
Reality is never as accurate, and is more noisy than a good simulation. Reality is only an approximation of the ideal.
 
  • #10
Baluncore said:
Reality is never as accurate, and is more noisy than a good simulation. Reality is only an approximation of the ideal.
A real experiment is a test of the experimenter, the theory and the equipment. A simulation is a test of the software and no more; A simulator won't show you where the theory may be wrong or (unless there's a table of likely errors, included) how dodgy some measuring setups can be.

A simulator is just a circular argument; very useful at some stages in a design. The problem is that people tend to favour simulations above an analytical solution (which somebody else had to work out in oder to wrote the simulator). That applies to massive cosmology simulators as much as SPICE but there are few alternatives to simulations to work out the formation of galaxies.
 
  • #11
sophiecentaur said:
The problem is that people tend to favour simulations above an analytical solution (which somebody else had to work out in order to write the simulator).
Only the most trivial analytic solutions are worth the time, and those I can see in my head. Bigger circuit analytic solutions are like crossword puzzles, you get the satisfaction of completing the solution, but they slow you down.

For real circuits, it is always quicker to build a SPICE model than solve it analytically. Once you have a model, you can optimise the circuit without needing to repeat the solution process.

As with software development, when a simulation produces a spurious result, minimising the circuit, while still exhibiting the problem, makes it possible to determine if there is a bug in the simulator, in your understanding of the simulator, or the circuit.

This thread began based on a minimised circuit that showed a spurious result. It was obvious that the sum of all currents to and from the neutral node must be zero, no analytic solution was needed for that. The question was, why did the simulator not agree with expectations. The answer was, as usual, operator inexperience. Experience will only rise if you climb the learning curve. Asking questions speeds the climb, and reduces the injuries.
 
  • #12
Baluncore said:
Turn the inductor's polarity dots on, and you will see that L1 is backwards in the circuit compared to L2 and L3. The inductor currents then total less than 1 fA, which for me, is close enough to zero.
The three pulse sources are identical, as is the neutral voltage.

Yes I saw they were the same. Didn't he want three-phase as in, 120° phase shift from each source?
 
  • #13
Baluncore said:
Turn the inductor's polarity dots on, and you will see that L1 is backwards in the circuit compared to L2 and L3. The inductor currents then total less than 1 fA, which for me, is close enough to zero.
The three pulse sources are identical, as is the neutral voltage.

Yes I saw they were the same. Didn't he want three-phase as in, 120° phase shift from each source?
 
  • #14
@boB_K7IQ
The schematic became 3PH later in Mosfetsim3.asc in post #5.

To make up for the lack of documentation, anyone prepared to build a SPICE model, deserves the assistance they request.
 
  • #15
Baluncore said:
Only the most trivial analytic solutions are worth the time, and those I can see in my head. Bigger circuit analytic solutions are like crossword puzzles, you get the satisfaction of completing the solution, but they slow you down.
Yes, but you've left out the third option which I'll call an analytic approximation. Which is an incredibly common (and powerful) engineering approach. So, if the approximations are reasonable, you can analyse (on paper) a multi-stage amplifier one stage at a time, for example. This is the most enlightening way to understanding circuits, IMO.

But yes, exact solutions may not be worth the effort for big, or highly coupled, networks. That's what simulators are really best at.

Finally, the reason you can "see in my head" the solution is because you've done the analytical solutions so many times before. That's not true for many others, yet. This is really why I dislike the "simulator first" approach, except for pros.
 
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  • #16
DaveE said:
Finally, the reason you can "see in my head" the solution is because you've done the analytical solutions so many times before.
That was what I was getting at. You can do endless simulations and get 'an answer' each time but simulations very seldom help with understanding the pattern. A dozen circuit simulations can't be guaranteed to give the message that the differential of the current waveform will give you the voltage waveform. Even accurate graphs that can be plotted numerically are not recognisable as sin(x)/x or exp ((a-x)2) unless someone who 'knows' the analysis has pointed it out first.
Simulations are brilliant for delivering results but are they really enough for getting a proper understanding of what goes on in Science.

The analysis (supported by countless careful measurements) tells us that the currents sum to zero. When it appears to fail, it's time to look for the fault in the input. (GIGO)
 

1. What is the concept of verifying the three currents add to 0?

The concept of verifying the three currents add to 0 is based on Kirchhoff's Current Law, which states that the sum of currents entering a node in an electrical circuit must equal the sum of currents leaving that node. This means that the algebraic sum of all currents at a given point in a circuit must be equal to 0.

2. Why is it important to verify the three currents add to 0?

It is important to verify the three currents add to 0 because it ensures that the conservation of charge is being upheld in an electrical circuit. If the sum of currents at a node does not equal 0, it indicates a potential error or malfunction in the circuit that needs to be addressed.

3. How do you verify the three currents add to 0 in a circuit?

To verify the three currents add to 0, you must first identify a node in the circuit where multiple currents converge. Then, using Kirchhoff's Current Law, you can write an equation stating that the sum of all currents entering the node must equal the sum of all currents leaving the node. By solving this equation, you can determine if the three currents add to 0 or not.

4. What are the consequences if the three currents do not add to 0?

If the three currents do not add to 0, it indicates that there is a violation of Kirchhoff's Current Law in the circuit. This could lead to inaccurate measurements, malfunctioning of devices, or even damage to the circuit components. It is important to identify and correct any errors in the circuit to ensure proper functioning.

5. Can the three currents ever add to a value other than 0?

No, according to Kirchhoff's Current Law, the three currents at a node must always add to 0. This is because charge is conserved in an electrical circuit, and any violation of this law would mean that charge is either created or destroyed, which is not possible. Therefore, the sum of three currents at a node must always be 0.

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