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I was in the lab and measured the analogue voltage at the output of an RC filter VF4 (top circuit). I now want to do simulations with these measurements to see how loading this voltage affects it. I can not use a source with zero output impedance, since this would not be realistic. If I apply Thevenins equivalent voltage source, I get the lower circuit with the measured voltage VF4 as a voltage source and parallel R4/C4 as source impedance. Does this conversion hold true for all frequencies? It should, but my intuition tells me something is wrong.

TLDR; does Thevenin apply to AC circuits as in the circuits below?

thank you very much

equivalent circuit?

Edit: Maybe rephrasing my question helps: Lets say I have measured a voltage VF4 in the EMC lab (the output of an RC filter as shown in the top circuit, measured with very high impedance so it's an unloaded "impedance divider"). I now would like to know the difference a second RC filter would make to that signal (with different time constant). For this I create a voltage source in tina with the measured voltage as .csv-input. If I now connect the second RC filter directly to that voltage source, my simulation results will be wrong, since the voltage source has zero internal resistance / impedance. I would like to know which internal impedance I should use for this voltage source to achieve accurate simulation results.

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    \$\begingroup\$ Very high frequencies would be fully present at VF5 because C5 would pass them along (bypassing R5.). But in the case of VF4 it would be close to 0 because C4 would short them to ground. So the two aren't equivalent. (Not sure if you think they are or if there's some other question I've missed.) \$\endgroup\$ Commented Aug 7 at 8:47
  • \$\begingroup\$ But very high frequencies are no longer present in the measurement of VF4 (which is used as an input on the second source), since the RC filter has already removed them prior to measuring. Or am I missing something? \$\endgroup\$ Commented Aug 7 at 8:53
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    \$\begingroup\$ I'm just speaking very generally. If the AC signal has a very low frequency then C4 won't short it out and C5 won't bypass R5, either. Then you just have R4 in one circuit and R5 in the other. So near DC they are about the same. It's only when the frequency increases that these two circuits are no longer similar. As far as your questions go, I can't say I'm getting into your mind very well. When I can put myself into someone else's thinking, then I can re-read their writing and find a match and a segue to help out. But I'm not able to put myself into yours. I'd know, if so. \$\endgroup\$ Commented Aug 7 at 9:04
  • \$\begingroup\$ Are R5 and C5 supposed to be the Thevenin equivalent impedance for R4 and C4? \$\endgroup\$ Commented Aug 7 at 14:46
  • \$\begingroup\$ @mathkeepsmebusy yes exactly \$\endgroup\$ Commented Aug 7 at 16:58

2 Answers 2

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My understanding (now) is that you want to compare the following two circuits under all load conditions and all frequencies.

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the magnitude (and phase) of voltage VF4 varies with frequency.

Picking an arbitrary load of 100 \$\Omega\$, we see that they have identical frequency responses.

magnitude phase

I now want to do simulations with these measurements to see how loading this voltage affects it.

You can easily simulate the above circuits (click the link) and modify the load. I have simulated with a capacitor in parallel with the load resistor, and the main result is the same, i.e. the two circuits have identical frequency response. Given that I chose an arbitrary resistor and an arbitrary capacitor, I am confident that for any load the circuits will be equivalent.

Again, note that the magnitude (and phase) of voltage VF4 varies with frequency, if a fixed AC voltage source is substituted for VF4 in the right hand side of circuit B, the circuits will be equivalent at a single frequency.

Lets say I have measured a voltage VF4 in the EMC lab (the output of an RC filter as shown in the top circuit, measured with very high impedance so it's an unloaded "impedance divider").

Are you assuming that the voltage you measure is the same for all frequencies?

I would like to know which internal impedance I should use for this voltage source to achieve accurate simulation results.

The Thevenin equivalent impedance you chose is correct. If you are getting simulation results that show a lack of equivalence it may be due to using a voltage source for VF4 which has a fixed magnitude.

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  • \$\begingroup\$ Thanks for your very extensive answer. Maybe I've phrased my question wrong. For thevenin's equivalent source you need an internal resistance (here internal impedance) and a equivalent voltage source. My question was if the two circuits produce the same load behaviour, when in simulation I add the measured voltage VF4 as an input voltage to the second circuits voltage source (in DC circuits, this would be a voltage divider with a load on it). Yes, about the frequency behaviour I'm fully aware, but if I insert a capacitor in simulation it already is an impedance in the spice equations, isntit? \$\endgroup\$ Commented Aug 7 at 13:51
  • \$\begingroup\$ in short: I only expect the two circuits I have shown above to be equivalent if a different voltage source is used, the second one having the unloaded output from the first one as a voltage source \$\endgroup\$ Commented Aug 7 at 14:02
  • \$\begingroup\$ this is exactly what I was looking for. In tina, I was not able to measure VF4 and use it as an voltage source with zero output impedance again as you did above. Thank you very much! \$\endgroup\$ Commented Aug 14 at 6:28
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Does this conversion hold true for all frequencies?

and

does Thevenin apply to AC circuits as in the circuits below?

Yes it totally does.

However, in practical circuits having non-zero dimensions, transmission line effects comes into play when signal wavelengths approach the physical size of conductors.

You might also consider skin effects in conductors. Skin effect lowers the effective conduction area of a wire or wound component thus, increasing net resistance. This also happens more at higher frequencies.

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  • \$\begingroup\$ Thank you very much, the frequencies concerned are well below transmission line effects. Thanks for the detailed answer \$\endgroup\$ Commented Aug 7 at 11:21
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    \$\begingroup\$ Andy, the two circuits are not equivalent at all frequencies in this case. \$\endgroup\$ Commented Aug 7 at 12:52
  • \$\begingroup\$ Does the op define the spectral characteristic of Vf4 in the diagram @MathKeepsMeBusy and anyway, I focussed on the questions and regard the diagram as unconnected to the questions. \$\endgroup\$ Commented Aug 7 at 13:01
  • \$\begingroup\$ I expect the source Vf4 to mimic the spectral response to voltage Vf4 in the upper diagram (for example). The op isn't that clear on this so maybe I'm wrong? \$\endgroup\$ Commented Aug 7 at 13:05
  • \$\begingroup\$ the input from the second circuit is the unloaded measured output from the first circuit, therefore higher frequency contents have already been removed by the RC filter \$\endgroup\$ Commented Aug 7 at 14:05

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