Simple circuit to measure discharge over time

For some reason, I can not plot any graph data, and to me the circuit looks like it should work, although it does not. I've put in each value, and used the same sizes for each component as what I did in the real test, but I'm lacking the diagram at the moment.

Could anybody point me in the correct direction? It's much appreciated!

by HarryTorry
March 13, 2012

Harry, There are a few problems I saw immediately:

  • No Ground connection
  • The interconnect lines run through the cap and Voltage source (this may be only a cosmetic error)
  • The switch never changes state - use the time dependent switch instead so that it will start in one condition and then change
  • The voltage you were putting across the polarized cap was backward - this is definitely not a problem for the simulation, as the simulator ignores the polarization direction.

I twiddled the bits around and got this to work fine: The label Vcap is just so the plotted data has a sensible name.

by CarlSawtell
March 13, 2012

Thanks a lot for this, I never ground my circuit, I don't know why not but I didn't. Maybe because the power supply did it?

I did this for screenshot for my assignment. I did have a ground connection, but removed it, since it wasn't (visibly, or to my knowledge) part of the circuit. When I originally did this, I did have a timed switch as well!

Thanks for this!

by HarryTorry
March 13, 2012

The ground is critical for the simulator, but in this case it can be put on any arbitrary node in the circuit. (The simulated result for the currents through all the components and the voltages across them will be the same wherever we put the ground.)

But depending on where we put the ground, when we calculate that the current voltage across the capacitor is 10V, my circuit would say the voltage at Vcap is -10V. Placing the ground at the bottom of "Power" would make the single-ended voltage read +5V. With the bottom of R1 grounded, it would read 0V. The simulator keeps track of the voltages at each node; without knowing which to reference, there is no way to solve for those voltages. (All nodes in an ungrounded schematic might be floating 20kV above ground for all the simulator can figure.)

The general rule is that there needs to be a DC solvable path from every node to ground. (Tricks are used in the simulator to allow some exceptions; for example nodes that lack a path due to a switch being open may still work)

by CarlSawtell
March 13, 2012

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