Newbie Needs Op Amp Help on Homework Problem

In summary, the conversation discusses the proper use of an ideal op-amp model in solving a homework problem involving KCL equations. The participants clarify that V+ and V- are equal in an ideal op-amp due to negative feedback, but this is not always the case in real op-amps. They also discuss the importance of considering the high gain and external feedback in analyzing op-amp circuits. The final solution to the homework problem is provided as V0 = 9V.
  • #1
DefaultName
180
0
I'm new to op-amps, and would like some guidance. I have a quick question on this HW problem. I'm not asking you to do it for me, but I want to see if my hunch is correct:

http://img66.imageshack.us/img66/8872/screenshot01bb7.jpg [Broken]

Now, using ideal op-amp model... I'm going to assume:

V+ = V1

V- = V1 because of neg. feedback?

Then apply KCL at the node where the R's meet?
So,

Vx / 1k + (Vx-V0) / 2K + (Vx-V2)/2k = 0
 
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  • #2
When you say "I'm going to call it V1" I hope you meant "the terminals of an ideal op amp are at the same voltage, so V+ = V- = V1"

If you didn't mean that, calling two different voltages V1 was not a good plan.

When you applied KCL, you forgot the current flowing in the other 2K resistor (between V2 and the opamp).
 
  • #3
AlephZero said:
When you say "I'm going to call it V1" I hope you meant "the terminals of an ideal op amp are at the same voltage, so V+ = V- = V1"

If you didn't mean that, calling two different voltages V1 was not a good plan.

When you applied KCL, you forgot the current flowing in the other 2K resistor (between V2 and the opamp).


Yes, that's what I meant.. sorry for the confusion. I've updated my KCL eq. in the first post -- does it look good?
 
  • #4
DefaultName said:
Yes, that's what I meant.. sorry for the confusion. I've updated my KCL eq. in the first post -- does it look good?

Yes, it looks OK now.

Now, using ideal op-amp model... I'm going to assume:
V+ = V1
V- = V1 because of neg. feedback?

The equations are right, but it is not because of "the neg. feedback", it's because an ideal op-amp always forces V+ = V- (whatever the rest of the circuit consists of).
 
  • #5
AlephZero said:
Yes, it looks OK now.



The equations are right, but it is not because of "the neg. feedback", it's because an ideal op-amp always forces V+ = V- (whatever the rest of the circuit consists of).

No, it is indeed because of the negative feedback. There is no internal mechanism for an opamp to hold its inputs together. It's the high internal gain and the external negative feedback that holds V- = V+. That's important to understand.
 
  • #6
Thanks guys...

I simplified it to 4Vx - V0 - V2 = 0...so,

V0 = 4Vx - V2, where Vx = V- = V+ = 7V? and V2 = 7V (because of the balance since it is an ideal op-amp)? I ended up gettin 4*7 - 7 = 21.0 V for V0.. it says I am incorrect.
 
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  • #7
berkeman said:
No, it is indeed because of the negative feedback. There is no internal mechanism for an opamp to hold its inputs together. It's the high internal gain and the external negative feedback that holds V- = V+. That's important to understand.

OK I understand that is true for real op amps and for designing practical circuits.

But isn't it true that for analysing a circuit with the ideal op amp model, then the assumptions are always i+ = i- = 0 and v+ = v- ?
 
  • #8
DefaultName said:
Thanks guys...

I simplified it to 4Vx - V0 - V2 = 0

Correct.

V0 = 4Vx - V2, where Vx = V- = V+ = 7V? and V2 = 7V (because of the balance since it is an ideal op-amp)? I ended up gettin 4*7 - 7 = 21.0 V for V0.. it says I am incorrect.

No. V1 is the same as V+ because they are joined by a wire link.
V1 = V+ = V- = 4V.

V2 is not the same as V-, because they are joined by the 2k resistor which has current flowing through it.
 
  • #9
AlephZero said:
Correct.
No. V1 is the same as V+ because they are joined by a wire link.
V1 = V+ = V- = 4V.

V2 is not the same as V-, because they are joined by the 2k resistor which has current flowing through it.

I see.. Since Vx is in the same node (shares the same wire) as the V-, Vx = V- = 4V.

So

(4 * 4) - 7 = 9 Volts, thanks guys!
 
  • #10
AlephZero said:
OK I understand that is true for real op amps and for designing practical circuits.

But isn't it true that for analysing a circuit with the ideal op amp model, then the assumptions are always i+ = i- = 0 and v+ = v- ?
Fair enough, but as soon as you are asked to do the calcs for transfer functions of real opamps, you have to take things like the finite (but high) gain and non-zero input currents (and offsets) and non-zero input voltage offsets into account. Also, if you connect up an opamp as a comparator with positive feedback to set the hysteresis, you will definitely not get V- = V+. I just wanted to be sure the OP kept that straight for future applications. o:)
 
  • #11
AlephZero said:
OK I understand that is true for real op amps and for designing practical circuits.

But isn't it true that for analysing a circuit with the ideal op amp model, then the assumptions are always i+ = i- = 0 and v+ = v- ?

no. if you do not have negative feedback, your op-amp becomes a comparator and saturates to very nearly the positive power supply voltage if v+ > v- and saturates to very nearly the negative power supply voltage if v+ < v- . since they are extremely high impedance inputs, v+ and v- are determined by analyzing the rest of the circuit surrounding those nodes.

actually the op-amp does not know and does not care if there is negative feedback or not, but, because of the extremely high gain, if it is in a negative feedback configuration, the output voltage will be forced to adjust to what it has to in order for v- to very nearly be equal to v+.
 
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1. What is an op amp?

An op amp is a type of electronic device that amplifies the voltage difference between its input terminals. It is often used in analog circuits to amplify and manipulate signals.

2. How does an op amp work?

An op amp typically has two input terminals, a non-inverting terminal and an inverting terminal, and one output terminal. The voltage difference between the input terminals is amplified and outputted through the output terminal. The gain or amplification factor of an op amp can be adjusted using external resistors.

3. What is the difference between an inverting and non-inverting op amp?

In an inverting op amp, the input signal is connected to the inverting terminal and the output is connected to the inverting terminal through a feedback resistor. In a non-inverting op amp, the input signal is connected to the non-inverting terminal and the output is connected to the inverting terminal through a feedback resistor.

4. How do I solve a problem involving op amps?

To solve a problem involving op amps, you will first need to understand the circuit and its components. Then, you can use circuit analysis techniques such as Kirchhoff's laws and Ohm's law to determine the output voltage or other parameters of the circuit.

5. What are some common applications of op amps?

Op amps have a wide range of applications such as audio amplifiers, voltage regulators, filters, and signal processing circuits. They are also commonly used in instrumentation and control systems.

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