Simulation aborted because there are errors during netlisting

I have built the following circuit:

http://i1226.photobucket.com/albums/ee410/jean28x/pspicemodel.jpg

However, when I try to simulate it, I get an error that says "Simulation aborted because there are errors during netlisting. Please refer to the sessions log."

What am I doing wrong? I'm new with PSPice so i'm not sure what's wrong.

Thanks.

Simon Bridge
Homework Helper
Did you refer to the sessions log and read the errors?

Where did you go wrong?

Mine ran just fine. You have selected 'ac analysis/noise' in simulation profile, right?

Attachments

• 29.7 KB Views: 2,163
• 49.8 KB Views: 1,954
Why you circuit don't work? The answer is simply you forgot about power supply. Unpowered opamps DO NOT WORK!

Where did you go wrong?

Mine ran just fine. You have selected 'ac analysis/noise' in simulation profile, right?

I hadn't selected the Noise in simulation profile. I select AC Sweep/Noise analysis and then which options to I select? The frequency es 10000 Hz so I know that but what do I put in the Noise analysis part?

Where did you go wrong?

Mine ran just fine. You have selected 'ac analysis/noise' in simulation profile, right?
I ran it while changing options in the noise area and got this mistake:

**** 10/22/12 13:20:02 *********** Evaluation PSpice (Nov 1999) **************

** Profile: "SCHEMATIC1-SimulationCircuitos2" [ C:\PsPice Stuff\projecto2circuitos2final-SCHEMATIC1-SimulationCircuitos2.sim ]

**** CIRCUIT DESCRIPTION

******************************************************************************

** Creating circuit file "projecto2circuitos2final-SCHEMATIC1-SimulationCircuitos2.sim.cir"
** WARNING: THIS AUTOMATICALLY GENERATED FILE MAY BE OVERWRITTEN BY SUBSEQUENT SIMULATIONS

*Libraries:
* Local Libraries :
* From [PSPICE NETLIST] section of pspiceev.ini file:
.lib "nom.lib"

*Analysis directives:
.AC DEC 100 10000 1000
------------------$ERROR -- Invalid value .NOISE 10 50 -------$
ERROR -- Must be independent source (I or V)
.PROBE
.INC "projecto2circuitos2final-SCHEMATIC1.net"

**** INCLUDING projecto2circuitos2final-SCHEMATIC1.net ****
* source PROJECTO2CIRCUITOS2FINAL
V_V1 N00036 GND DC 0Vdc AC 6.250Vac
R_R1 N00036 N00042 200k
R_R2 N00042 N00076 50k
R_R3 N00042 N00140 40k
C_C1 N00076 N00140 10n
C_C2 GND N00042 25n
X_U1A N00076 GND N00140 GND N00140 LM324
V_V2 N00113 GND 9Vdc
V_V3 GND M_UN0001 9Vdc

**** RESUMING projecto2circuitos2final-SCHEMATIC1-SimulationCircuitos2.sim.cir ****
.INC "projecto2circuitos2final-SCHEMATIC1.als"

**** INCLUDING projecto2circuitos2final-SCHEMATIC1.als ****
.ALIASES
V_V1 V1(+=N00036 -=GND )
R_R1 R1(1=N00036 2=N00042 )
R_R2 R2(1=N00042 2=N00076 )
R_R3 R3(1=N00042 2=N00140 )
C_C1 C1(1=N00076 2=N00140 )
C_C2 C2(1=GND 2=N00042 )
X_U1A U1A(+=N00076 -=GND V+=N00140 V-=GND OUT=N00140 )
V_V2 V2(+=N00113 -=GND )
V_V3 V3(+=GND -=M_UN0001 )
_ _(GND=GND)
_ _(GND=GND)
.ENDALIASES

**** RESUMING projecto2circuitos2final-SCHEMATIC1-SimulationCircuitos2.sim.cir ****
.END

You must be kidding me.
Your circuit will work if you add symmetric power supply to op amp. Positive terminal to pin 4. Negative to pin 11 and GND to ground.

Simon Bridge
Homework Helper
You must be kidding me.
Your circuit will work if you add symmetric power supply to op amp. Positive terminal to pin 4. Negative to pin 11 and GND to ground.
It usually helps someone follow your advice if you show how you came to the conclusions you did. What made you realize that the power supply was not hooked up correctly?

Simon Bridge
Homework Helper
Code:
*Analysis directives:
[b].AC DEC 100 10000 1000[/b]
------------------$ERROR -- Invalid value [b].NOISE 10 50[/b] -------$
ERROR -- Must be independent source (I or V)
What do the three numbers in the .AC directive mean?

It usually helps someone follow your advice if you show how you came to the conclusions you did. What made you realize that the power supply was not hooked up correctly?
Just one look at the diagram

And what you see ?? No power supply.

Where did you go wrong?

Mine ran just fine. You have selected 'ac analysis/noise' in simulation profile, right?
Are you sure?? Output voltage is only 42μV. Typical example of GIGO in the case of simulators, garbage in=sophisticated garbage out.

Attachments

• 20.7 KB Views: 7,340
Last edited:
Simon Bridge
Homework Helper
Could this have something to do with the second error message?

davenn
Gold Member
Could this have something to do with the second error message?
what the 6.25V AC ? thats the input signal me thinks

D

Simon Bridge
Homework Helper
what the 6.25V AC ? thats the input signal me thinks
This is what I'm talking about! Now what role has OP got it playing in the circuit diagram?

Probably best to continue this part of the discussion in private since it is about how to answer questions effectively and not actually about OPs problem.

I hadn't selected the Noise in simulation profile. I select AC Sweep/Noise analysis and then which options to I select? The frequency es 10000 Hz so I know that but what do I put in the Noise analysis part?
You dont necessarily have to put noise analysis in your simulation if that is not something you want to do.

Code:
.AC DEC 100 10000 1000
------------------$ERROR -- Invalid value .NOISE 10 50 -------$
ERROR -- Must be independent source (I or V)
The ".AC" command syntax is
Code:
.AC <sweep type> <points value> <start frequency value> <end frequency value>
In your case start freq is greater than end freq. You have to set it correctly in simulation window.

The ".NOISE" command syntax is
Code:
.NOISE V(<node> [,<node>]) <name> [interval value]
You did not give V(output node) and name(noise source) etc.

I'm telling you again, you probably do not need noise analysis. So turn it off. AC analysis will run just fine without noise analysis.

You must be kidding me.
Your circuit will work if you add symmetric power supply to op amp. Positive terminal to pin 4. Negative to pin 11 and GND to ground.
Jony, I know very well that the circuit will not work hooked up like that. But that does not mean the circuit will not simulate and give out the garbage value. The issue I tried to solve was with the simulation procedure. Once that is done Jean can see whether his circuit produces desired result or not, and come to us again.

vk6kro
I tried a LTSpice simulation of this circuit.

Because of the positive feedback, the output depends on the initial voltage on the input, but it always goes to one supply rail or the other.

Assuming this is a mistake, I reversed the inputs to the opamp and it then behaves like a low pass filter with a 6dB cutoff of about 300 Hz.

As a matter of interest, if the 10 nF capacitor is replaced with a smaller value, the circuit becomes a bandpass filter. For example, a 220 pF capacitor in this position produces a peak output at about 1480 Hz.

Last edited:
As a matter of interest, if the 10 nF capacitor is replaced with a smaller value, the circuit becomes a bandpass filter. For example, a 220 pF capacitor in this position produces a peak output at about 1480 Hz.
This is not true. This circuit (correct diagram) is a multiple feedback low-pass filter.
And if you change capacitor to 220pF the only think that has change except cut-off frequency is a quality factor. Q has change from 0.7 to Q = 4.767 for 220pF but it is still a low pass filter.

Attachments

• 4.7 KB Views: 1,079
• 2.5 KB Views: 1,096
vk6kro