Solving Ideal Op Amp Circuits with Power Sources

AI Thread Summary
To solve an ideal op amp circuit with independent power sources, the output voltage can be determined using Kirchhoff's Voltage Law, resulting in a calculated output of +2V. An ideal op amp maintains a zero voltage difference between its inputs, adjusting the output to achieve this balance. The discussion emphasizes that the circuit's behavior is consistent only under ideal conditions, and real components may yield different results. It is suggested to consult additional resources for a deeper understanding of op amp behavior. Overall, grasping the fundamental principles of op amps is crucial for accurate circuit analysis.
Sinister
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Quick question: how do you solve an ideal op amp circuit with the two terminals connected to independant power sources?
 
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I have no idea what you mean? Are you talking about the Vcc and Vee pins of the opamp or +/-inputs?
 
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Something like this


*SORRY if this has turned into a homework question, >.<*
 
The output should be +2V. reason is the definition of an ideal opamp is it has infinite open loop gain of any voltage between the +ve and -ve input. In a closed loop feedback configuration, the output of the opamp will do whatever it takes to make the differential voltage between +ve and -ve input to be zero. You have a +1V source from the output to the -ve input, you have a +3V from ground to the +ve input. So the ideal will do whatever to make the -ve input to go to +3V which is +2V at the output. The 2K resistor has no effect with the circuit. This is basically a voltage follower with a twist of the 1V source which will give an offset of -1V to whatever you put at the +ve input.

Is this a trick question?
 
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correct, but a brief explination on how ?
 
Hehe, sorry but my textbook only gives a half page explanation on OP Amps, so its not very good.
but thanks for that!
 
I basically gave you the explanation for that. The definition of an opamp is it has a +ve and -ve input and an output. The differential gain ( of whatever presented BETWEEN the + and - input ) is very very high meaning under normal operation condition, the voltage difference between the two inputs is always zero. If anything try to put a voltage across the inputs, the output of the opamp will try to compensate by driving in direction to eliminate the voltage difference at the two input. That is exactly what I explained. Read it through and thing about it. Also Yahoo opamp and read a little more. Then if you have more question, come back and I'll try to help more.
 
Always - the opamp will do its best to keep the two inputs ar same voltage.

When you're just beginning, write Kirchoff's Voltage Law from one input to the other and set to zero


starting at +, walking around to - and equating to zero;
3v - Vo - 1v = 0
Vo = 2

sanity check - does Vo lie between the power supply rails? If not something is wrong.

that should get you started. Soon you'll be like Yungman - read them as easy as resistor color codes.

old jim

You might look for the old books we used in 70's -
Don Lancaster's books (opamp cookbook?)
National Semioconductor catalog and AN 31

and Texas Instruments "OpAmps for Everyone" which you should download from TI.com and print and bind to pass on to your grandkids.
 
jim hardy said:
Always - the opamp will do its best to keep the two inputs ar same voltage.

When you're just beginning, write Kirchoff's Voltage Law from one input to the other and set to zero


starting at +, walking around to - and equating to zero;
3v - Vo - 1v = 0
Vo = 2

sanity check - does Vo lie between the power supply rails? If not something is wrong.

that should get you started. Soon you'll be like Yungman - read them as easy as resistor color codes.

old jim

You might look for the old books we used in 70's -
Don Lancaster's books (opamp cookbook?)
National Semioconductor catalog and AN 31

and Texas Instruments "OpAmps for Everyone" which you should download from TI.com and print and bind to pass on to your grandkids.

Does this only works if one input is connected to the ground and the other is connected to the output
 
  • #10
Sinister said:
Does this only works if one input is connected to the ground and the other is connected to the output

If the + input at ground and the output connect to -input. If the opamp want to do whatever it takes to bring the differential voltage to 0V, what do you think the voltage would be.

Say if you initially has output of opamp at +1V and you put 0V at the +input:

1) What is the differential voltage?

2) With the differential voltage, what direction the output going to swing?

3) If the output swing toward that direction, when is it going to stop?

Think about this.

One can write 20 pages on the close loop behavior of an opamp, you really need to get some material to read if you want to understand it.
 
  • #11
Your circuit doesn't make any sense for ideal voltage sources and an ideal op amp.

Your question is similar to asking "I have two equations,
x = 1 and x = 3, and I want to solve them for x". There is no solution, because the equations are inconsistent iwth each other.

Of course if you built the circuit from real components, the real components are NEVER ideal, so you would get some output voltage that you could measure. But you can't predict what will happen by assuming the components are ideal. Different types of op-amp would most likely give different outputs for the same input.
 
  • #12
I thought the OPs first post and the schematic posted clearly explain it. Yungman gave the correct answer.
 
  • #13
"Does this only works if one input is connected to the ground and the other is connected to the output"

nope. Try it out a few times.
Then you'll begin to see that Yungman gave you excellent answers.

Aleph , you might want to revisit your analysis...
 
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