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In order to better understand RLC behavior under current sources, I constructed the simple circuit below and attempted to analyze it. However, my equations do not agree with simulation results.

The equations are simple, and use standard calculus, yet they don't seem to produce correct results.

enter image description here

My analysis is in some ways consistent with simulation but in some ways gets very different results. Is my analysis correct? If so, why does it not agree with simulation?

I've already asked about this on an electronics site. While they confirmed that simulation does not agree with the equations, no one was able to spot any error in them or suggest more correct ones. So I decided I needed to ask here.

Analysis shows: $$ V_{out} = I_{src} \omega^2 LCR_{load}$$ since

$$ I_{src} = I_0 e^{j \omega t} \\ V_{L1} = V_1 - L \frac{dI_{src}}{dt} \\ = V_1 -jI_{src} \omega L$$ and $$ I_{load} = C \frac{dV_{L1}}{dt} \\ = \omega^2 L C I_{src}$$ so $$V_{out} = I_{load}R_{load} \\ = I_{src} \omega^2 LCR_{load}. $$

This suggests that $V_{out}$ is linear in $L, C, R$ and quadratic in $f$ and should be in phase with $I_{src}$. However, when I simulate it, I find that similar relationships, but not exact, and that $V_{out}$ hits a limit at a few MHz after which increasing $f$ has no effect. Why?

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    $\begingroup$ You seem to be after the AC output voltage, but are including a DC voltage source in your analysis, why? I also don't see how you came up with $V_{L1} = V_1 - L \frac{dI_{src}}{dt}$. In any case, shorting the voltage source and straightforward AC analysis of the remaining impedance network quickly reveal that $$V_{out}=\frac{\omega^2RLC}{1+j\omega RC-\omega^2RLC}I_{src}.$$ $\endgroup$
    – Puk
    Commented May 31 at 15:48
  • $\begingroup$ @Puk: The voltage drop across $L1$ is $L\frac{dI}{dt}$, and the voltage at the point $L1$ touches the capacitor is $V_1 - \text{ drop across } L1$. Could you explain where you got that denominator from? (Your numerator is identical to mine, but my equations show a denominator of simply $1$). Why does it not show in my equations? $\endgroup$
    – SRobertJames
    Commented May 31 at 15:54
  • $\begingroup$ 1: $V_{L1}$ would typically indicate the voltage drop across the inductor. Please label these voltages on your diagram, or define them clearly in your question. 2: The current through the inductor is not $I_{src}$, so the equation still doesn't work. 3: Just remove the voltage supply (short it) and analyze the simple impedance network with two branches that remains. I can't do the analysis for you step-by-step as that would go against the purpose/rules of this site. $\endgroup$
    – Puk
    Commented May 31 at 16:02
  • $\begingroup$ Consider also the fact that as $\omega \to \infty$, the inductor is open and the capacitor is short, and the output voltage is simply $-RI_{src}$. It makes no sense for it to keep increasing with frequency without bound. $\endgroup$
    – Puk
    Commented May 31 at 16:09
  • $\begingroup$ While they confirmed that simulation does not agree with the equations, no one [on EE site] was able to spot any error in them or suggest more correct ones. That is not entirely true as there are four valid responses. $\endgroup$
    – Farcher
    Commented Jun 1 at 8:33

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