Kirchoff's Laws Series-Parallel Question

MikeMl · Jan 26, 2016

LTSpice follows the conventions established by Berkley Spice, PSpice, and other Spices... The manual page you linked is for PSpice.

Spice's choice of current direction thru a source enables Spice to show power delivered by the source as negative power, while dissipative circuit elements show power consumed as positive power. This was already discussed by Ghar in this linked thread: https://forum.allaboutcircuits.com/t...nt-always-measured-negative-in-ltspice.40854/

I covered flipping resistors end-for-end (so that the sign of the current through the resistor is reversed) in post #3 of this linked thread:
https://www.electro-tech-online.com/threads/ltspice-current-probe.109196/.

Here is a simple sim that demonstrates Spice's power convention.

V1 is fixed at 5V, while V2 varies from 0V to 10V in small steps. The x-axis is V(V2) = V(b). The upper plot pane shows the current through V1 and V2 (per Spices's definition as discussed in the linked pdf). The traces were added to the upper plot pane by using the "amprobe" cursor and clicking on the voltage sources; the plotting is done automatically. Note that for different values of V2's voltage, at times the current in both sources and the resistor reverses...

The lower plot shows LTSpice's view of the power contributed by or dissipated by both sources and the resistor. The plots were obtained automatically by clicking on V1, V2 and R1 with the "thermometer" (power) cursor displayed. LTSpice shows if a source is contributing power or dissipating power seamlessly. Positive power is always dissipation. Note that the resistor only dissipates power. Note that all power goes to zero when V(V2)=V(V1). Note that when V(V2)=10, the power dissipated by both the resistor and V1 is 5W each.

How would you have done it? (Hindsight is always 20:20 )

Ratchit · Jan 26, 2016

MikeMl said:
LTSpice follows the conventions established by Berkley Spice, PSpice, and other Spices... The manual page you linked is for PSpice.

Correct, every simulation program I know of is backward.

Spice's choice of current direction thru a source enables Spice to show power delivered by the source as negative power, while dissipative circuit elements show power consumed as positive power. This was already discussed by Ghar in this linked thread: https://forum.allaboutcircuits.com/t...nt-always-measured-negative-in-ltspice.40854/

Yes, Ghar did mention that. But it makes more sense to have a voltage or current source that provides power to have a positive designation. After all, increasing power output should be represented with a larger number, not a lessor number like a negative designation does.

I covered flipping resistors end-for-end (so that the sign of the current through the resistor is reversed) in post #3 of this linked thread:
https://www.electro-tech-online.com/threads/ltspice-current-probe.109196/.

If I reverse the placement of a resistor in the real world, the voltage polarity and current direction do not change. Neither should the resistors in circuit simulators.

Here is a simple sim that demonstrates Spice's power convention.

View attachment 97098

V1 is fixed at 5V, while V2 varies from 0V to 10V in small steps. The x-axis is V(V2) = V(b). The upper plot pane shows the current through V1 and V2 (per Spices's definition as discussed in the linked pdf). The traces were added to the upper plot pane by using the "amprobe" cursor and clicking on the voltage sources; the plotting is done automatically. Note that for different values of V2's voltage, at times the current in both sources and the resistor reverses...

The lower plot shows LTSpice's view of the power contributed by or dissipated by both sources and the resistor. The plots were obtained automatically by clicking on V1, V2 and R1 with the "thermometer" (power) cursor displayed. LTSpice shows if a source is contributing power or dissipating power seamlessly. Positive power is always dissipation. Note that the resistor only dissipates power. Note that all power goes to zero when V(V2)=V(V1). Note that when V(V2)=10, the power dissipated by both the resistor and V1 is 5W each.

As others have pointed out, the current designation from a voltage source is opposite what a ammeter would indicate in the real world. The power dissipated by a resistor will be the same no matter what the current direction.

How would you have done it? (Hindsight is always 20:20 )

I would have it like I do with pencil and paper. Using mathematical convention, I would have designate current leaving the positive terminal of a voltage source as positive. The power it supplies would also be I*V which would be positive. A resistor would be a positive current times a negative voltage which would make the resistor power negative. This means that the resistor absorbs electrical energy and dissipates it as heat.

Ratch

MikeMl · Jan 26, 2016

Ratchit said:
...
Correct, every simulation program I know of is backward.

Only to your way of thinking...

...I would have it like I do with pencil and paper. ... A resistor would be a positive current times a negative voltage which would make the resistor power negative. This means that the resistor absorbs electrical energy and dissipates it as heat.

Ratch

So according to you, power delivered by a source is I*E, while power dissipated in a resistor is -E*I; how is that an improvement??? I like my way (Spice's way) better. I have yet to go to a store and buy a resistor that is rated to dissipate -2W.

So according to you, if current in the resistor is positive if it flows from left to right, to get a negative voltage across the resistor, the potential at the left end of the resistor would have to be lower than the potential at the right end. You are espousing electron current!

You completely glossed over the issue that sources can sometime supply power, sometimes dissipate power, which requires a current reversal.

The only thing that matters is that whatever convention you choose, you apply it consistently, and rigorously.

Ratchit · Jan 26, 2016

MikeMl said:
Only to your way of thinking...

So according to you, power delivered by a source is I*E, while power dissipated in a resistor is -E*I; how is that an improvement??? I like my way (Spice's way) better. I have yet to go to a store and buy a resistor that is rated to dissipate -2W.

So according to you, if current in the resistor is positive if it flows from left to right, to get a negative voltage across the resistor, the potential at the left end of the resistor would have to be lower than the potential at the right end. You are espousing electron current!

You completely glossed over the issue that sources can sometime supply power, sometimes dissipate power, which requires a current reversal.

The only thing that matters is that whatever convention you choose, you apply it consistently, and rigorously.

Ratchit · Jan 26, 2016

MikeMl said:
Only to your way of thinking...

I gave a good reason why my suggestion is better than what is now in place. You should address that suggestion instead of stating the obvious.

So according to you, power delivered by a source is I*E, while power dissipated in a resistor is -E*I; how is that an improvement???

Because it makes more sense to designate losses as a negative quantity.

I like my way (Spice's way) better.

To each his own.

I have yet to go to a store and buy a resistor that is rated to dissipate -2W.

I have yet to go to a store and buy a power supply that supplies -100 watts.

So according to you, if current in the resistor is positive if it flows from left to right, to get a negative voltage across the resistor, the potential at the left end of the resistor would have to be lower than the potential at the right end. You are espousing electron current!

How do you reason that? If an ammeter indicates the current direction is from left to right, and I put the negative terminal of the voltmeter on the left side and the positive terminal on the right side of the resistor, the voltmeter will indicate a negative value according to the mathematical convention. You need to think over what you aver above.

You completely glossed over the issue that sources can sometime supply power, sometimes dissipate power, which requires a current reversal.

If a power supply provides current in one direction it is stands to reeason that forcing a reverse current will cause the power output to reverse. Why is such an obvious thing worthy of explanation?

The only thing that matters is that whatever convention you choose, you apply it consistently, and rigorously.

No, what matters is that the convention is correct in the first place, so that false results do not propagate.

Ratch

MikeMl · Jan 26, 2016

Let me ask you, if I connect the red lead of my voltmeter to the left end of a resistor, and the black lead to the right end of a resistor, and I see a positive reading, which way is the current flowing? left to right? right to left?

Ratchit · Jan 26, 2016

MikeMl said:
Let me ask you, if I connect the red lead of my voltmeter to the left end of a resistor, and the black lead to the right end of a resistor, and I see a positive reading, which way is the current flowing? left to right? right to left?

You are showing a more positive voltage on the left side of the resistor. Therefore, the mathematical current direction is from left to right through the resistor.

Ratch

MikeMl · Jan 27, 2016

Ratchit said:
You are showing a more positive voltage on the left side of the resistor. Therefore, the mathematical current direction is from left to right through the resistor...

Is the "mathemetical current direction" a signed quantity? If so, in your world, which current direction carries what sign?

Ratchit · Jan 27, 2016

MikeMl said:
Is the "mathemetical current direction" a signed quantity? If so, in your world, which current direction carries what sign?

Every number is a signed quantity, whether explicitly designated or implied. So what does that question mean? Let me iterate the mathematical current convention as it relates to a component. If current, as determined by a ammeter, leaves the more positive terminal of a device, that device is supplying power to the circuit. If the current is entering the more positive terminal of a device, then the device is taking power from the circuit. For instance, if a resistor measures -4 volts and -6 volts on either side, the current will enter the more positive -4 volts side and the resistor will be considered a power sink. A voltage or current source that shows a current departing the more positive side will be considered a power source. This convention will be in compliance the mathematical current convention and Tellegen's theorem, which declares that the power entering a circuit equals the power leaving the circuit. LTSpice does not comply with that convention.

Ratch

MikeMl · Jan 28, 2016

Ratchit said:
... Tellegen's theorem, which declares that the power entering a circuit equals the power leaving the circuit.

I find it ironic that a proof of Tellegen's theorem defines the current direction through the voltage source the same way that LTSpice does. Note that the author of this page choose I1 to be equal but opposite to I2. B is a node of the circuit. By KCL, the sum of the currents into that node have to equal zero.
LTSpice's calculations are consistent with KCL.

Also, Tellegen's theorem states that the summation of instantaneous powers for the n number of branches in an electrical network is zero. It says nothing about dissipation carrying either a plus or minus sign. LTSpice's calculation's show that power in equals power out...

Ratchit · Jan 28, 2016

MikeMl said:
I find it ironic that a proof of Tellegen's theorem defines the current direction through the voltage source the same way that LTSpice does.

First of all, that link shows a verification of Tellegen's theorem, not a proof of it. I would think that the proof of it would reference the Conservation of Energy Principle. The direction of current around a circuit has nothing to do with a proof. As long as the currents and voltages are consistent, the sum of the energy sources and sinks will equal zero.

Note that the author of this page choose I1 to be equal but opposite to I2. B is a node of the circuit. By KCL, the sum of the currents into that node have to equal zero.

Which is inconsistent with the real world, isn't it? I would have done it the right way by keeping the currents in the same direction and changing the voltage reading across the source the way it really happens.

LTSpice's calculations are consistent with KCL.

No they are not. LTSpice shows current entering the positive terminal of a voltage source. That's a no-no for the mathematical current convention. Because it does not square with what a ammeter would indicate, others have noticed this anomaly and commented on it. The "excuse" given for this is that it makes the power calculations negative when the source gives power to the circuit. This is supposed to balance the loss of power in the resistors which is considered positive. In post #25 of this thread I showed why this is wrong thinking.

Also, Tellegen's theorem states that the summation of instantaneous powers for the n number of branches in an electrical network is zero. It says nothing about dissipation carrying either a plus or minus sign. LTSpice's calculation's show that power in equals power out...

Other verifications do show dissipations as one sign and sources as the opposite sign. The bookkeeping method has nothing to do with Tellegen's theorem. In LTSpice, power in equals power out provided you hook up the components with the correct node designations.

View attachment 97112

MikeMl · Jan 28, 2016

Ratchit said:
... As long as the currents and voltages are consistent, the sum of the energy sources and sinks will equal zero.

Which, if you bothered to follow the example above, is exactly what LTSpice shows...

...I would have done it the right way by keeping the currents in the same direction and changing the voltage reading across the source the way it really happens.

Which, as just I showed in #30, is exactly what LTSpice does. How could it be otherwise if the summation of currents into node a is zero per KCL?

The voltage reading is correct as shown. How could it be otherwise if the sum of voltages around the loop adds to zero, per KVL? What is your "Right Way"???

LTSpice shows current entering the positive terminal of a voltage source.

Look again! It shows a negative current entering the positive terminal of the voltage source V1. To me, that means the current is actually flowing out of the positive terminal... So where is the inconsistency???

Other verifications do show dissipations as one sign and sources as the opposite sign. The bookkeeping method has nothing to do with Tellegen's theorem. In LTSpice, power in equals power out provided you hook up the components with the correct node designations.

Which is what I have been saying all along.

I think all of your rants about Spice in general, and LTSpice in particular just fell apart, with the exception of the pin swapping on two-terminal components.

The solution is really simple. Mark pin 1 on the two-terminal, symmetrical components, and then insert them into the schematic so that positive current flows into pin 1. You might not know that until you actually do a simulation, and actually look to see which direction current is actually flowing. Note that this problem occurs only with a graphics schematic editor. In text-based Spice, you always had to define a-priori "which-way-around" a two-pin component is instantiated.

Note that the power calculation done by LTSpice is correct regardless of which way around a symmetrical component is drafted in a schematic.

Hopelessly Pragmatic...

Ratchit · Jan 28, 2016

MikeMl said:
Which, if you bothered to follow the example above, is exactly what LTSpice shows...

I never said the energy sum of the sink and sources were not zero in LTSpice. I said that energy losses should have been designated as minus, not plus. Just flipping the signs is not going to change their sum to a nonzero value.

Which, as just I showed in #30, is exactly what LTSpice does. How could it be otherwise if the summation of currents into node a is zero per KCL?

That is not what the documentation I submitted in the second link of post #20 says. To quote: "N+ is the positive node and N- is the negative node, as shown in Fig. 4-5(a). Positive current flows from node N+, through the voltage source, to the negative N-." Your run shows negative current entering N+ (according to the arrow) and exiting at N-. The current now runs in the correct direction, but you must have switched leads of the voltage source in LTSpice to do that. In any case, what should be is that the current arrow of I(V1)( should point upwards and the current be a positive value.

The voltage reading is correct as shown. How could it be otherwise if the sum of voltages around the loop adds to zero, per KVL? What is your "Right Way"???

You are confusing your LTSpice run with the verification of the Tellegen verification link you submitted. That verification did not use LTSpice, so what does your question mean?

Look again! It shows a negative current entering the positive terminal of the voltage source V1. To me, that means the current is actually flowing out of the positive terminal... So where is the inconsistency???

There is no inconsistency in the circuit value results, except for designating losses as positive values. The problem is that you have to negate the current value to accommodate the wrong direction, which should not be necessary.

Which is what I have been saying all along.

You were criticizing my using plus and minus to show energy gains and losses. I am glad you agree that the bookkeeping method is irrelevant to Tellegen's theorem.

I think all of your rants about Spice in general, and LTSpice in particular just fell apart, with the exception of the pin swapping on two-terminal components.

Think what you want, but others have noticed the same thing. LTSpice still puts out the wrong signs for energy losses and gains, and still has to use a kluge to get the currents moving in the right direction. Therefore my complaints have not fallen apart.

The solution is really simple. Mark pin 1 on the two-terminal, symmetrical components, and then insert them into the schematic so that positive current flows into pin 1. You might not know that until you actually do a simulation, and actually look to see which direction current is actually flowing. Note that this problem occurs only with a graphics schematic editor. In text-based Spice, you always had to define a-priori "which-way-around" a two-pin component is instantiated.

That will not cure the opposite signs of energy gains and losses. And it looks "funny" to see a voltage or current source pointing the wrong way with a negative value.

Note that the power calculation done by LTSpice is correct regardless of which way around a symmetrical component is drafted in a schematic.

As it should be, but the sign will be wrong.

Hopelessly Pragmatic...

Hopelessly pedantic and dogmatic

Ratch

MikeMl · Jan 28, 2016

As usual, you contribute nothing useful, and just want to argue, forever, for arguments stake. I have better things to do with my time...

Ratchit · Jan 28, 2016

MikeMl said:
As usual, you contribute nothing useful, and just want to argue, forever, for arguments stake. I have better things to do with my time...

I do not support LTSpice, so I can't fix its problems. But they should be exposed and explained as others have done.

Ratch

Kirchoff's Laws Series-Parallel Question

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