PSPICE Help: Integrating Op-Amp

Thread Starter

random_digit

Joined Apr 21, 2007
8
Hello,

I'm working on a problem involving designing an ac integrator driven by a 10-V p-p square wave and using Probe to compare waveforms.

I've drawn out the integrator but for some reason it would not create the waveforms for me and said my circuit had errors. Specifically, "Less than 2 connections at node $N_0004" :confused:

I looked over my schematic and it looks fine to me, it matches up with what the integrators look like in my textbook. If I had to take a guess to were it would be wrong, it would be either my VPULSE or the non-inverting input of the op-amp.

Could somebody please help me pinpoint my problem, I'm pretty stumped. I've attached my circuit below. Thank you, I appreciate it.
--
Arthur
 

Attachments

Papabravo

Joined Feb 24, 2006
13,544
It is impossible to see where node $N_0004 is located on the schematic. In the netlist generated from the schematic is a node which is connected to only one point. You need to find and eliminate that node. How you do that is PSPICE implementation dependent.
 

Thread Starter

random_digit

Joined Apr 21, 2007
8
Well, I went over it again with a friend who is having the same problem. He found out that it is something with the non-inverting input.

So if I were to follow what you said Papabravo, I am to just delete this? Connect straight to ground?

I'm still stuck -- it won't draw anything out for me. Please help, thank you!
 

JoeJester

Joined Apr 26, 2005
4,259
What signal are you using as the input? I know you said it was 10 volts pk-pk. Is that 5 volts plus and 5 volts minus? Is that 10 volts plus, 0 volts or the reverse of that?

What is the frequency of your 10 volt pk-pk signal?
 

Thread Starter

random_digit

Joined Apr 21, 2007
8
I'm using a VPULSE part, going from -5V to 5V.

The problem has 2 parts, one simulation at 1 kHz and the other at 200 Hz. So under the part properties for the VPULSE I put 1/1000 for the PER, but to no avail. I can kind of guess what the waves are going to look like, but I need to have these waveforms present and printed out.

Thank you for all the help so far!
 

Thread Starter

random_digit

Joined Apr 21, 2007
8
* From [PSPICE NETLIST] section of pspiceev.ini:
.lib "nom.lib"

.INC "chap11.37.net"

**** INCLUDING chap11.37.net ****
* Schematics Netlist *



R_R1 $N_0002 $N_0001 20k
R_Rf $N_0001 $N_0003 50k
C_C $N_0001 $N_0003 .1uF
V_V2 $N_0002 0
+PULSE -5V 5V 1ns 500ps 500ps 50ns .001s
E_U1 $N_0003 0 VALUE {LIMIT(V($N_0004,$N_0001)*1E6,-15V,+15V)}
R_Rc 0 $N_0004 14.2857k

**** RESUMING chap11.37.cir ****
.INC "chap11.37.als"



**** INCLUDING chap11.37.als ****
* Schematics Aliases *

.ALIASES
R_R1 R1(1=$N_0002 2=$N_0001 )
R_Rf Rf(1=$N_0001 2=$N_0003 )
C_C C(1=$N_0001 2=$N_0003 )
V_V2 V2(+=$N_0002 -=0 )
E_U1 U1(OUT=$N_0003 +=$N_0004 -=$N_0001 )
R_Rc Rc(1=0 2=$N_0004 )
.ENDALIASES


**** RESUMING chap11.37.cir ****
.probe


.END

ERROR -- Less than 2 connections at node $N_0004
 

n9352527

Joined Oct 14, 2005
1,198
Somehow in your netlist U1 is connected only to two nodes $N_0003 and 0 (ground), although your alias seems to understand this and is correct. U1 actually needs to be connected to three nodes, $N_0003, $N_0004 and $N0001.

Check your schematic diagram, make sure all the nodes and wires are really connected and not just very near or on top of each other. Create a new netlist to confirm this.
 

JoeJester

Joined Apr 26, 2005
4,259
+PULSE -5V 5V 1ns 500ps 500ps 50ns .001s
I have questions about this. Your setting for a pulse width of 50 nS? I believe this exercise calls for a square wave, 50% duty cycle, at the two frequencies of interest.

This could be the reason for your "no output" condition, expecially if your attempting to view it over n milliseconds.
 

Thread Starter

random_digit

Joined Apr 21, 2007
8
Well, I figured out how to fix my problem with it not showing an output -- since the netlist said that I had less than 2 connections at n_0004 (my non-inverting input), I put a huge resistor in parallel with Rc so it creates a voltage divider, but yet keeps the value of Rc mostly intact. Doing this resulted in an output, which has me estatic even though I'm still trying to fiddle with it to get a nice looking output (damn vpulse).

I think the reason it had me to do this had something to do with the nature of an operating amp and its inherent 'virtual ground' between the inverting and non-inverting inputs.

But oh well, thanks for everything you guys!

Cheers.
 
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