Symbol | Examples |
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1. For DC analysis a constant 1V voltage source, and for transient
analysis a 5V 50% dutycycle 1MHz square wave, rising at 100ns with a rise/fall
time of 10ns:
value: 1V PULSE(0V 5V 100ns 10ns 10ns 490ns 1us) 2. For DC analysis a constant 5V voltage source, and for transient analysis
an undamped 1V (peak to peak) sine of 100KHz, starting at time 0:
3. For DC and transient analysis a constant 5V voltage source, and for
AC analysis a 1V (peak to peak amplitude) source:
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1. A constant current source of 5 Amps both for DC and transient analysis:
value: 5A 2. For DC analysis a constant 5A source, and for AC analysis a 2A (peak
to peak amplitude) source with 90 degs phase shift:
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These are the transient source formulas:
PULSE(t) = v0 @ 0+n*per .. td+tr+n*per,
v1 @ td+tr+n*per .. td+tr+pw+tf+n*per,
v0 @ td+tr+pw+tf+n*per
.. 0+(n+1)*per, n: 0, 1, 2, 3,...
SIN(t)= vo @ 0 .. td, vo + va exp(-(t-td) theta) sin(2 pi freq (t + td)) @ td .. end
EXP(t) = v0 @ 0 .. td1, v0 + (v1
- v0)*(1 - exp(-(t - td1)) / tau1)) @ td1 ..
td2,
v0 + (v1 - v0)*(1 - exp(-(t - td1))
/ tau1)) - (v1 - v0)*(1 - exp(-(t - td2)) /
tau2))
@ td2 .. end
PWL(t) = v0 @ 0, v1 @ t1, .., vn @ tn .. end
SSFM(t) = vo + va sin(2 pi fc t + mdi sin(2 pi fs t))
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