Find Vout as a function of V1 and V2 (OP Amps)

In summary: Alternatively, change your V1 value to something that won't drive the op amps to the rails. Try V1 = 0.5 V for example.
  • #1
rugerts
153
11
Homework Statement
Find an expression for the output voltage as a function of the input voltages
Relevant Equations
KVL, KCL, ideal op amp assumptions
Here's the circuit in question:
1573932365241.png

Solution:
1573932390559.png
1573932409810.png


Now, when I try simulating in LTSpice, this is what I get:
1573932481696.png


So, Vout appears to be around -13 V, which doesn't agree with the equation if V1=V2= 5 is plugged in.

Does anyone see the mistake here?
THanks.
 
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  • #2
Your final LTSpice circuit seems to have an input polarity error for U3.

You didn't state it in your problem statement (I may have missed it), but V1=V2=5V? I see that at the end of your post. I don't see those numbers in your handwritten calculations (but again I may be not understanding).

Can you unambiguously draw the circuit and label each node with the voltage? Thanks.
 
  • #3
Ponder what constraints the Vcc supplies for the op amps imply, and also what real world constraints there are for LM741's maximum output swing.

From your LTSpice diagram it seems that the Vcc voltages for the opamps are ±15 V. That's fine. However, if you look at the first stage, that 5 V input voltage is going to want to push 1 mA into the 5 kΩ input resistor. That 1 mA will want to drop 15 V across the 15 kΩ feedback resistor. That's the full rail potential as set by Vcc.

A real life 741 opamp can't achieve a full rail swing (the opamp output can only approach to within about a volt of the Vcc voltages at best, and often closer to one and a half or more volts). That's why your simulation shows -13.9 V for Vb.

The next stage has a theoretical gain of -40/10 = 4, but with the input already at -15 V (well, -13.9 for "real life" parts) it clearly can't perform that operation, and its output must swing as close as it can to the appropriate rail. So another 13.9 V result.

I'll leave it there and encourage you to take a fresh look at the final stage using the concepts introduced above.
 
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  • #4
In my reply above I assumed that you were working with "real world" components rather than ideal op amps, since you chose specific components in your simulation. If the op amps are instead considered ideal with no constraints on the output voltage swing, clearly the analysis would go in a different direction since the achievable output swings for the op amps would no longer be a constraint to impose.
 
  • #5
@berkeman
@gneill

So, for the hand calculations, I'm doing them for any V1 or V2 since my goal is to find an expression for Vout as a function of V1 and V2.
For the LTSpice simulation, I just plugged in 5V for both sources and 15V for the OP Amps. There's no particular reason for these number choices, I sort of just assumed they would work.

In my hand calculations, there is the assumption of the ideal op amp. In the simulated LTSpice circuit, I think it does just a good approximation of the ideal state.

Good point about the polarities. That's sort of confusing me. I tried changing them around but it's still not getting me the result I obtained doing the hand calculation.
 
  • #6
The calculation (handwritten) is correct - however, the polarity for the last opamp (LT simulation) is wrong (as mentioned already by berkeman)
 
  • #7
LvW said:
The calculation (handwritten) is correct - however, the polarity for the last opamp (LT simulation) is wrong (as mentioned already by berkeman)
It seems like it's still wrong.
1574013375238.png

1574013949697.png

Based on the hand calculation, I'm expecting Vout = (36*5-5)/2 = 87.5 V
 
Last edited:
  • #8
rugerts said:
Based on the hand calculation, I'm expecting Vout = (36*5-5)/2 = 87.5 V
With 741 op amps running with Vcc = 15 V? That'll never happen. See post #3.

Try replacing the 741's in your simulation with the generic "opamp" component. No Vcc specification required.
 
  • #9
gneill said:
With 741 op amps running with Vcc = 15 V? That'll never happen. See post #3.

Try replacing the 741's in your simulation with the generic "opamp" component. No Vcc specification required.
How so? We used 741 OP Amps in lab running on 15V. Unless you mean for this specific configuration shown? Also, in the original schematic up top shown it says 741 next to each op amp.
 
  • #10
rugerts said:
How so? We used 741 OP Amps in lab running on 15V. Also, in the original schematic up top shown it says 741 next to each op amp.
You are wondering why your hand calculations are not matching the simulation. The explanation is in post #3. If you want the simulation to match your calculations, replace the "real world" 741's with generic opamps that don't have any "real world" limitations.

Alternatively, change your V1 value to something that won't drive the op amps to the rails. Try V1 = 0.5 V for example.
 
  • #11
gneill said:
You are wondering why your hand calculations are not matching the simulation. The explanation is in post #3. If you want the simulation to match your calculations, replace the "real world" 741's with generic opamps that don't have any "real world" limitations.

Alternatively, change your V1 value to something that won't drive the op amps to the rails. Try V1 = 0.5 V for example.
Understood and thank you. Simulation works and agrees with hand calculation.
 
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FAQ: Find Vout as a function of V1 and V2 (OP Amps)

1. What is an OP Amp?

An OP Amp (Operational Amplifier) is an electronic device that amplifies the difference between two input signals and outputs the result as a single output signal. It is commonly used in electronic circuits to perform mathematical operations such as addition, subtraction, differentiation, and integration.

2. How do I find Vout as a function of V1 and V2 using an OP Amp?

To find Vout as a function of V1 and V2 using an OP Amp, you can use the following formula: Vout = A(V2 - V1), where A is the amplification factor of the OP Amp. This formula assumes that the OP Amp is ideal, meaning it has infinite input impedance, zero output impedance, and infinite open-loop gain.

3. What are the limitations of using an OP Amp to find Vout as a function of V1 and V2?

Some limitations of using an OP Amp to find Vout as a function of V1 and V2 include the fact that it is only accurate for ideal OP Amps and may not hold true for non-ideal OP Amps. Additionally, the formula assumes that the OP Amp is being used in negative feedback mode, meaning the output is connected to the inverting input. If the OP Amp is used in positive feedback mode, the formula will not hold true.

4. Can I use multiple OP Amps to find Vout as a function of V1 and V2?

Yes, you can use multiple OP Amps to find Vout as a function of V1 and V2. In this case, you would need to connect the output of one OP Amp to the input of the next, creating a chain of amplifiers. The final output would then be the result of all the amplifiers in the chain.

5. Are there any practical applications of finding Vout as a function of V1 and V2 using OP Amps?

Yes, there are many practical applications of using OP Amps to find Vout as a function of V1 and V2. Some common examples include audio amplifiers, voltage regulators, and signal filters. OP Amps are also used in a variety of instrumentation and control systems, as well as in mathematical and scientific equipment.

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