Andrew Beckett
Senior Fellow
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Life, don't talk to me about Life...
Posts: 1742
Bracknell, UK
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If you set the write parameter on the transient analysis to write out the DC solution at the beginning of the transient (with rforce at the default value), you'll see that it shows:
# CHECKPOINT_VERSION 1 # Generated by spectre (mode: Spectre) from circuit file `dg1.scs' during analys is tran. # 10:37:49 AM, Thur Feb 23, 2017 # Number of equations = 3 L0:1 1 osc 0
What is going on is that the inductor is a short at DC (and the capacitor is open). The initial condition is implemented as the norton equivalent of the desired voltage on the capacitor in series with 1 ohm (the rforce value). This is 1 amp in parallel with 1 ohm. So that's why you end up with 1Amp through the inductor.
Now, because at the resonant frequency, the impedance is 100 ohms, that will (V=IR) generate 100V at the resonant frequency.
If you set rforce=100, the norton equivalent is 0.01A with 100ohm in parallel to give 1V across the inductor. Again, because it's shorted at DC, you just get 0.01A through the inductor. So the checkpoint file contains:
# CHECKPOINT_VERSION 1 # Generated by spectre (mode: Spectre) from circuit file `dg1.scs' during analys is tran. # 11:26:30 AM, Thur Feb 23, 2017 # Number of equations = 3 L0:1 0.01 osc 0
Again, V=IR at the resonant frequency would then result in 1V at the resonant frequency (which is what you want).
Regards,
Andrew.
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