# Help with operational amplifier

Discussion in 'Homework Help' started by esmeco, Oct 15, 2006.

1. ### esmeco Thread Starter New Member

Jul 10, 2006
9
0
I'm trying to simulate this ciruit on microsim schematics:

http://i75.photobucket.com/albums/i2.../amplifier.jpg

This's what I've got so far:

http://i75.photobucket.com/albums/i2...simcircuit.jpg

What I want to see is the final graph with the values of V0,V1,V2.V1 is 1V and V2 varies from -5V to 7V.I'm not sure if my circuit is totally right so I'd really appreciate if someone could correct it where it is wrong, or give me some pointers on how to correct it.

Just one more question,how do I read the values of the error |V0 experimental - V0 theoric|?Why do the values from v0 theoric differ so much from those of V0 experimental(from class)?Shouldn't the V0 theoric values stand between +15vcc and -15vcc?This's the table with the experimental and theoric results value(didn't put the simulated values because I'm not sure if the microsim circuit is correct):

http://i75.photobucket.com/albums/i2...sultstable.jpg

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
None of the links you provided appear to work properly for me.

Can you click on them and successfully access images?

hgmjr

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
143
Likewise, it claims that all the linked pages cannot be found.

Dave

Jul 10, 2006
9
0
5. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I would be very interested in learning where you got the figures you have on the table with the heading "Theoretical Vo".

Even the best rail-to-rail opamp cannot exceed its power supply rails. That means that the output of your opamp cannot output a voltage greater than +15V nor less than -15V.

While you are at it, maybe you could also tell us how you obtained your "Experimental V0" figures.

For example, when V1 = 1 and V2 = 3, I get an output of 6V. In this case, the voltage at the positive terminal is 2. That means that the voltage at the negative terminal must also be 2 volts. With V1 = 1 volt, then 1 volt is dropped across the 1K resistor R1. That means that 1 milliamp is flowing from the junction of R1 and R2 toward the V1 terminal. Since the opamp's input impedance is very high, then the only source for this 1 milliamp of current must be coming from the current flowing in through R2. To get 1 milliamp of current flowing into the junction of R1 and R2 from R2 then the voltage at the output of the opamp must be 4 volts more positive than the voltage at the opamp's negative terminal. That would put the opamp's output terminal at 6 volts. Your table indicates the theoretical output voltage to be 11 volts.

hgmjr