1
\$\begingroup\$

I tested a simple circuit with AC voltage source and, when I probed the source input for voltage and current, I was surprised to find that they were 180 degrees out of phase.

I tested in the LTSpice simulator.

Is it normally like that in real world or something to do with the simulator or my circuit?

enter image description here

and the wave generated is

enter image description here

\$\endgroup\$
12
  • 1
    \$\begingroup\$ In the real world with a real resistive load voltage and current are in phase. \$\endgroup\$
    – Andy aka
    Oct 1, 2015 at 14:51
  • \$\begingroup\$ It can, use ICE, current leads voltage in a capacitor by 90 degrees and ELI where current lags voltage by 90 degrees in inductors. This is why AC analysis uses phasers. However, I am not sure why there is a shift in the circuit you show above. \$\endgroup\$ Oct 1, 2015 at 14:51
  • \$\begingroup\$ @JarrodChristman could u help me with the jargons - ICE , ELI \$\endgroup\$
    – aj_blk
    Oct 1, 2015 at 14:54
  • 1
    \$\begingroup\$ No it is not an issue of LTspice, all simulators behave like this. It is expected behaviour. He plotted the current through the voltage source, that current is negative when the current outside the source is positive. I have seen that on all simulators I worked with. \$\endgroup\$ Oct 1, 2015 at 15:06
  • 2
    \$\begingroup\$ Please add a labelled node/net to your circuit and plot that. Otherwise we have no idea what you're measuring. \$\endgroup\$ Oct 1, 2015 at 15:11

2 Answers 2

4
\$\begingroup\$

This is simply due to the passive sign convention used by LTSpice's current probe for the voltage source. By this convention, power sources such as voltage sources have negative power dissipation, which means that the source's current is negative when the voltage is positive and vice versa.

Notice the arrow on the current probe icon is pointing downward for this voltage source:

enter image description here

It also points downward for \$R_1\$:

enter image description here

The plotted currents are out of phase:

enter image description here

But obviously they are the same current, as in this circuit there is only one current path. They are just out of phase because they are taken in the opposite direction, as indicated by the probe icon.

\$\endgroup\$
4
2
\$\begingroup\$

The answer is simple, you plotted I(V1) instead of -I(V1) (or the reverse !) There is no phase shift, 180 degrees is also an inversion so I becomes -I.

Solution: where did you probe the current ? At the plus node of the resistor ? Then probe it at the minus node ! Or the reverse !

edit: Maybe LT spice only allows you to plot the current through the source, this current will be negative when the current coming out of the + node is positive. This is expected behaviour. All simulators do this.

Nothing fancy happening here :-)

\$\endgroup\$
10
  • \$\begingroup\$ no, i plotted the I passing through V1 thats the voltage source \$\endgroup\$
    – aj_blk
    Oct 1, 2015 at 14:56
  • \$\begingroup\$ Then plot the current through R1, it does not matter, it is the same current. Also a positive current coming out of the voltage source means that the current is flowing from - to + INSIDE the voltage source. That's what the simulator plots for you, the current INSIDE the voltage source. That current is negative when the current going out of the voltage source is positive. You are confused by the direction of the current. \$\endgroup\$ Oct 1, 2015 at 14:58
  • \$\begingroup\$ Why the -1 ?? If you have a better explanation please let me know. No reason to -1 my answer when you don't understand how the simulator works. \$\endgroup\$ Oct 1, 2015 at 15:14
  • \$\begingroup\$ I wasn't the one who downvoted but note that he clearly measured the current through the voltage source. The real question is where did he measure the voltage (n001). There are two nodes (nets) in this circuit (of non-zero voltage) because the voltage source is not directly connected to ground but through another resistor! \$\endgroup\$ Oct 1, 2015 at 15:20
  • 1
    \$\begingroup\$ The resistor is odd indeed, but in this case it does not matter as no current can flow to/from ground as there is only one connection. I guess the voltage was probed between the two lines in the middle. The voltage is as expected though so no issue. Sources with negative currents, that is what confuses people ! \$\endgroup\$ Oct 1, 2015 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.