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A tran simulation problem of RLC resonated network.

xxgenerall
Hi everyone,

A RLC resonant network is as Figure 1, seen at my post below.

L=1uH, C=1uF, R=100kOhms.

Source generates a sine wave signal. The source signal's frequency is 159.155kHz and its amplitude is 10V.

1. Theoretical analysis.

 The resonant frequency of LC network is about 159.155kHz. Under this frequency, the LC network's AC impedance is very large, and reaches about 1.5 MOhms. So if the circuit becomes steady, Vout is about 0V, whereas Vin-Vout is almost equal to Vin.

2. Tran simulation result.

A tran simulation is implemented with Cadence Spectre. The "stop time" is set to be 1s, so that the circuit can get steady. The tran simulation result is as Figure 2, seen at my post below.

It is seen that, Vout is almost equal to Vin.

By the way, sp and ac simulations are also carried out, which all indicate the LC network has a very large impedance at 159.155kHz.

The tran simulation result is so different from the theoretial analysis, and is not consistent with sp and ac simulation results. Why?

-----------------------------------------

Hi Mr. Beckett, Thanks for your help.
Parents

  •  Because the Q is so high, the lack of timesteps in a default simulation means that it will not be able to properly see the effect of the resonance.

    By forcing more timesteps, you can see the amplitude decreasing (but it's rather slow, because this will clearly have a long time constant).

    I simulated this netlist:

    //
    V0 (Vin 0) vsource type=sine ampl=10 freq=159.154943K mag=10
    L1 (Vin Vout) inductor l=1u
    C1 (Vin Vout) capacitor c=1u
    R0 (Vout 0) resistor r=100K

    //ac ac start=159.154K stop=159.156K step=.001
    ac ac start=155K stop=160K step=.001

    tran tran stop=1 method=traponly maxstep=1/(2000*159K)

    That number of timesteps is probably excessive - and I didn't wait for the simulation to end. If you look at the AC response, you can see that there's a very high Q on the resonsance. Looking at the beginning of the simulation (78million timepoints so far) shows the attached curve (obviously you can see the envelope).

    In real life there would be some non-idealities in there...

    Andrew.

Reply

  •  Because the Q is so high, the lack of timesteps in a default simulation means that it will not be able to properly see the effect of the resonance.

    By forcing more timesteps, you can see the amplitude decreasing (but it's rather slow, because this will clearly have a long time constant).

    I simulated this netlist:

    //
    V0 (Vin 0) vsource type=sine ampl=10 freq=159.154943K mag=10
    L1 (Vin Vout) inductor l=1u
    C1 (Vin Vout) capacitor c=1u
    R0 (Vout 0) resistor r=100K

    //ac ac start=159.154K stop=159.156K step=.001
    ac ac start=155K stop=160K step=.001

    tran tran stop=1 method=traponly maxstep=1/(2000*159K)

    That number of timesteps is probably excessive - and I didn't wait for the simulation to end. If you look at the AC response, you can see that there's a very high Q on the resonsance. Looking at the beginning of the simulation (78million timepoints so far) shows the attached curve (obviously you can see the envelope).

    In real life there would be some non-idealities in there...

    Andrew.

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