sir/mam,
i am trying to simulate an AC circuit which is attached in the below link:
i need to find phase angle of Is and I2
for finding Is i have attached a very low series resistance ( Rs ) (approximately equal to 0) as shown in the attached image link:
but the pspice program is not giving the desied results!!
i have copied my Pspice .CIR file below:
To find phase angle of current in AC circuit
Vs 1 0 AC 50
Rs 1 2 1e-30
R1 2 3 10
L1 3 0 0.1
R2 2 4 20
C1 4 0 1000u
.AC LIN 1 15.9 15.9
.print AC IP(Rs) IP(R2)
.end
output:
FREQ IP(Rs) IP(R2)
1.590E+01 -9.000E+01 2.659E+01
But the actual value should be:
IP(Rs) = -18.44 degree
IP(I2) = 26.59 degree
thus i am getting an incorrect value of IP(Rs) in Pspice!!
so please help me and guide me to simulate the circuit properly in Pspice
with thanks
Copyright © 2020 Cadence Design Systems, Inc. All rights reserved.
Try an actual sweep:
.AC LIN 100 15.0 15.99
Call the library:
.LIB
Open the Probe Window
.Probe
With these additional command in the CIR file, I get the expected results. In the PSpice Simulation Profile, just change the Sweep Parameters, the route for Capture will add the .LIB and .Probe commands.
still i am getting the same results!!!
eg.:
IP(Rs)
-9.000E+01
for all Freq !!!!