0
\$\begingroup\$

enter image description here

Why is AC current inversely proportional to voltage?

I think that current should be proportional to voltage, because according to \$V=IR\$, when \$R\$ is fixed, the bigger \$I\$ is, the bigger \$V\$ should be. But according to this simulation, the relationship between \$I\$ and \$V\$ is negative. Is the software the problem? I use Virtuoso Cadence to simulate the circuit.

I have another question. The direction of the arrow should be the direction of the current. Why is the first half wave negative?

Is the meter "flipped around" in 1 (below)? Or, should I modify something in 2? I had tried simulating 1, but it didn’t change the relation between the voltage and current.

enter image description here

\$\endgroup\$
  • 3
    \$\begingroup\$ Tried flipping the meter around? \$\endgroup\$ – user253751 Apr 13 '18 at 3:14
  • 2
    \$\begingroup\$ @XM551 flip either the voltmeter or the ammeter around. One but not both. This is just a polarity or sign convention error. No big deal. \$\endgroup\$ – mkeith Apr 13 '18 at 3:40
  • 1
    \$\begingroup\$ @XM551 What do you think it means when an ammeter shows a negative value? It means the current is going backwards through the meter. If you flip it around the current would be going forwards through the meter. Same for voltmeters. \$\endgroup\$ – user253751 Apr 13 '18 at 4:02
  • 5
    \$\begingroup\$ That's not inverse proportionality. Inverse proportionality is $$I \propto \frac{1}{V}$$. What you see is $$I \propto -V$$ \$\endgroup\$ – Ben Voigt Apr 13 '18 at 7:04
  • 1
    \$\begingroup\$ @XM551: No proportional means that both variables have a fixed (constant) ratio; even if the ratio is negative; What you should have written in your question is "Why has current the opposite sign as expected?". I is proportional to V, even if the sign is "the wrong way". \$\endgroup\$ – Curd Apr 13 '18 at 14:46
3
\$\begingroup\$

Cadence has a convention to use the current flowing out of the device as negative. You are measuring the current at the positive node of the current source so the current is negative. Measure it at the node connected to resistor, you will get positive current.

\$\endgroup\$
1
\$\begingroup\$

A flipped Ammeter while simulation - Probably what is happening in there.

enter image description here

Link to the simulation

\$\endgroup\$
  • \$\begingroup\$ Are you simulate the same point? \$\endgroup\$ – XM551 Apr 13 '18 at 12:14
  • \$\begingroup\$ Same ckt. Same measuring points. \$\endgroup\$ – Mitu Raj Apr 13 '18 at 12:23
  • \$\begingroup\$ I had modify my question,now i think the thing i can't understand is the meaning of "flipping the meter around",can you please see my question again?thx \$\endgroup\$ – XM551 Apr 13 '18 at 13:00
  • \$\begingroup\$ What is "V" in your graph ? Which point is your probe at ? Left or right side of Resistor ? \$\endgroup\$ – Mitu Raj Apr 13 '18 at 13:18
  • \$\begingroup\$ V is the voltage on the red line,the point is the green circle.you can see it from the first picture \$\endgroup\$ – XM551 Apr 13 '18 at 13:20
0
\$\begingroup\$

I think you are the confusing the word “meter” with your current source. They are two different things.

Changing the direction of your current source won’t fix your problem, if the direction of the measurement changes with it.

Is your plot measuring the flow of electrons or (conventionally) holes? When you add the current plot to your simulation, your CAD is likely adding it in a fix, albeit backwards, direction.

\$\endgroup\$
  • \$\begingroup\$ How do i know my plot is measuring the flow of electrons or (conventionally) holes? \$\endgroup\$ – XM551 Apr 13 '18 at 14:15
  • \$\begingroup\$ It is measuring conventional current. How do you know this? Because nobody every measures electrons. Ever. It is always conventional current. Just forget about electrons unless you are studying device physics or chemistry. \$\endgroup\$ – mkeith Apr 13 '18 at 16:37
  • 2
    \$\begingroup\$ @mkeith I agree, and just threw it out there to be sure. The only reason I considered it is this: my son has a class in electronics (middle school) and came to me for help earlier this week. I was surprised to learn that the students were all being taught their circuit analysis in electron-flow current! I was appalled, and seriously considered scheduling an appointment to talk to his teacher. \$\endgroup\$ – Blair Fonville Apr 13 '18 at 16:44
  • 1
    \$\begingroup\$ Wow, @BlairFonville, I too am appalled and surprised! \$\endgroup\$ – mkeith Apr 13 '18 at 16:49
  • 2
    \$\begingroup\$ @mkeith I was actually quite frustrated about it; especially so because I know he took the class to make me proud, and so I hide my disappointment. I really should talk to the teacher. \$\endgroup\$ – Blair Fonville Apr 13 '18 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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