achim.graupner
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Dear shuguang,
Description: Set up a schematic with a voltage source connected to the varactor. Set AC Magnitude to "1". Set up an AC-Simulation, sweep DC Voltage of the voltage source. Use 1/2Pi=0.159 as frequency. Plot the current (AC current) of the voltage source = capacitance value. When using any other frequency than above, divide the current by 2Pi and the frequency.
Explanation: AC computes small signal values, i.e. derivatives of the plotted signal vs. the AC sources. If plot the varactor's current IF("/V1/PLUS") the AC Analysis yields dIC/dVC. dIC/dVC = j*ω*C * vc = j*2πf*C * vc where vc is the small signal voltage (AC Magnitude), vc=1. Thus: C = dIC/dVC / 2πf / vc = IF("/V1/PLUS") / VF("/V1") / 2πf
Understandable? If not - please ask. Regards, Achim
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