# Question with multisim circuit

Discussion in 'General Electronics Chat' started by uofmx12, Aug 20, 2011.

1. ### uofmx12 Thread Starter Member

Mar 8, 2011
55
0
I have a question dealing with a circuit I am to construct in multisim. The first link is the small part of the circuit I am not quite sure on. And the second link is my circuit and what I constructed thinking of what would be correct. I used a potentiometer with +-10V supplies on each side, which I am not sure if it is correct. If I did wrong, can someone correct me please.

- What I am to construct:
http://i812.photobucket.com/albums/zz41/uofmx12/E21ckt.jpg

-What I constructed:
http://i812.photobucket.com/albums/zz41/uofmx12/E21.jpg

2. ### Potato Pudding Well-Known Member

Jun 11, 2010
684
92
You seem to still need a linear OPAMP powered by +10 and -10 Volt supplies.

Also you have 100k vs 10k expected potentiometer.

For compensation the potentiometer is meant to be a trimmer and it only makes certain that the circuits function is centered on the desired voltage.

Proportions of R3(R2 in your MultiSim) to R4(the potentiometer) will determine the tuning sensitivity.

3. ### uofmx12 Thread Starter Member

Mar 8, 2011
55
0
Yea, I saw the 100k to 10k for R4 mistake. For the Opamp, do I need a 5T opamp to have +-10V attached?

My main question was dealing with what was circled. I was not sure what the diagram meant when it had the arrow pointed to the resistor R4. I assumed it was a potentiometer but was not sure it my assumption was right, or if it was just a wire connected to the resistor.

4. ### Potato Pudding Well-Known Member

Jun 11, 2010
684
92
That circled part is definitely a potentiometer.

You need an OPAMP made for double sided supply with greater than 10Volts to the rails other than that I think you are probably fine. Some of that is only if you are trying to exactly duplicate the circuit. You sound like you might be just experimenting.

If this is homework then the instructor might have indicated what OPAMP model to use.

5. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
You have your V1 and V2 connected incorrectly. The only output that the pot R4 will have is -10v.

I usually start off by placing battery or voltage supply symbols on the left of the schematic; if I'm using a dual rail supply, I'll stack them one on top of the other, and place a ground between them. That way it's pretty obvious what's going on.

6. ### uofmx12 Thread Starter Member

Mar 8, 2011
55
0
No, nothing was mentioned. I am reading from a book. It says, assemble the circuit in the first figure using +-10V supplies. So I am assuming something like:
http://i812.photobucket.com/albums/zz41/uofmx12/E21ckt_1.jpg

From what I have gathered in this thread so far. With those specifications I am trying to make Vc(node C, purely DC) = 0V by adjusting R4 and leaving Node A open. (looking at the first link up above ckt). But using this ckt I posted in this post, when I adjust R4 the VDC in the voltmeter does not change from 9.33VDC.

7. ### uofmx12 Thread Starter Member

Mar 8, 2011
55
0
Ok, so in other words, V2 needs to leave and just leave the V1 there at -10V? I did that and connected R2 to the R4 resistor. Then trying to accomplish what I mentioned in the previous post with adjusting R4 to make Vc = 0V using the DVM, I used a 5M resistor and that took it from the 10k=9.33V to 5M=1.7V. Is is supposed to be that high of a resistor to make Vc = 0V?

8. ### SgtWookie Expert

Jul 17, 2007
22,183
1,728
No, you need to get the polarities correct at the ends of the pot.
It's easier if you have the voltage sources off by themselves.

It's too late for me to comment much more; I've had a long day.

9. ### uofmx12 Thread Starter Member

Mar 8, 2011
55
0
Sorry, Im not quite following. I see you said ' You have your V1 and V2 connected incorrectly. The only output that the pot R4 will have is -10v.' but obviously what I thought it meant was incorrect. Could someone else give input? Thanks