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TP Pspice

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toufik bendib
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0% found this document useful (0 votes)
31 views5 pages

TP Pspice

Uploaded by

toufik bendib
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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BENZAROUR date :1/11/2023

NOUZHA
4eme annee u_electronique

Practical work 1 : pspice simulation of fundamental

circuit
1/the netlist describing the circuit :

.R1 1 0 800

.R2 1 2 500

.R3 1 3 1K

.C1 3 0 33n

.I1 0 1 DC 4mA

.V1 2 0 DC 30V

2/The value of voltages V(1) and V(3) :

.R1 1 0 800

.R2 1 2 500

.R3 1 3 1K

.C1 3 0 33n

.I1 0 1 DC 4mA

.V1 2 0 DC 30V

.op

.end

V(1)=19.620v

V(2)= 19.620v

3/perform a continuous simulation :

.R1 1 0 800

.R2 1 2 500

.R3 1 3 1K

.C1 3 0 33n

.I1 0 1 DC 4mA

.V1 2 0 DC 30V
.DC V1 0 30 0.5

.probe

.op

.end

4/the forms of the voltages v(1) and v(3) as a function of v1:

V(1) = V(3) = ax + b (x=V1)

a= (5.84 - 3.84 )/(7.51 - 3.59) = 0.51 b= 1.23

V(1)=V(3)=0.51V1+1.23

5/ by using the command .TF :

.R1 1 0 800

.R2 1 2 500

.R3 1 3 1K

.C1 3 0 33n

.I1 0 1 DC 4mA

.V1 2 0 DC 30V

.TF V(3) v1

.op

.end

Req: 1.300E+03

Veq: 19.6920

6/perform a transient similation using .TRAN

.R1 1 0 800

.R2 1 2 500

.R3 1 3 1K

.C1 3 0 33n IC=0


.I1 0 1 DC 4mA

.V1 2 0 DC 30V

.op

.tran 1u 0.8m

.probe

.end

7/plot the simulated curve and find the voltageans the constant of capacity charge :

Final voltage: Vs=19.692V

. 0.63*Vs= 12.4V

by viewing the corresponding time for Vs= 12.4, we find that τ=43.22us

8/PULSE:

For Frequence= 1KHz

Vth= 1 0 PULSE (0 5 1f 1f 0.5m 1m)

C1 0 2 1300

Rth 2 1 1300

.tran 50u 4m

.probe

.end
Frequence= 10KHz

Vth= 1 0 PULSE (0 5 1f 1f 0.05m 0.1m)

C1 0 2 33n IC=0

Rth 2 1 1300

.tran 1u 0.4m

.probe

.end

Frequence =20KHz

Vth= 1 0 PULSE (0 5 1f 1f 25u 50u)

C1 0 2 33n IC=0

Rth 2 1 1300

.tran 0.5u 200m

.probe

.end
9/

10/plot the transfer function curves en DB and phase .AC :

11/The cutoff frequency:

The cutoff frequency is the frequency corresponding to G(dB)=-3dB and also Δfi(°)=-45° which is:
f=3.7kHz

12/the conclusion:

we observe, we identified that the implemented circuit functions as a low-pass filter. This means it
allows low frequencies to pass through unchanged while attenuating or cutting off higher
frequencies. The specific cutoff frequency for this filter was determined to be 3.7kHz. Consequently,
any signal with a frequency below 3.7kHz remains intact, while frequencies exceeding this value are
effectively filtered out or removed.

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