Find the gain in a non-ideal op amp - check my logic

In summary: If the gain is less than 1.0, then the voltage at the output will be less than the input voltage. So when you solve for V_{0}/V_{s}, you'll get a number that's less than 1.0. In summary, the op amp is designed to give a gain of 1.0, so the voltage at the output will be equal to the input voltage.
  • #1
jkuhling
1
0
To start I have attached an image of the circuit. We are to relate the [itex]V_{0}[/itex]to [itex]V_{s}[/itex]. I am not exactly sure where to start with it. I know that [itex]V_{-}[/itex] and the [itex]V_{+}[/itex] on the op amp have to have the same voltage. The resistor that bridges the two inputs is throwing me off.

Not sure if this is the right way to start but if I look at the current through the 400[itex]\Omega[/itex] resistor as [itex]I_{0}[/itex]. Then using KCL:

[itex]I_{0}[/itex]=[itex]I_{1}[/itex]+[itex]I_{2}[/itex].

Sense there is no current running through the op amp, then what ever the current is in the first 1200[itex]\Omega[/itex] is the same current in the second 1200[itex]\Omega[/itex]. By this the current at [itex]V_{0}[/itex] to be the current through the 600[itex]\Omega[/itex] resistor.

If I put in the voltage across the head minus the voltage in the tail stuff for the currents and then solve to the [itex]V_{0}[/itex]/[itex]V_{s}[/itex] should be the right way.

I am curious if I am thinking about this right or I am completely off my rocker.
Thanks in advance.
 

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  • #2
Just write the usual node or loop equations (I'm a node man myself).

I don't understand what you mean by "head" and "tail" stuff. Are we flipping coins here? :rolleyes:

Why is your post titled "non-ideal" op amp?
 
  • #3
jkuhling said:
The resistor that bridges the two inputs is throwing me off.
Don't let it. The presence of that resistor should pique your interest! :smile:
Sense there is no current running through the op amp, then what ever the current is in the first 1200[itex]\Omega[/itex] is the same current in the second 1200[itex]\Omega[/itex].
Correct. So with that pair of 1200Ω resistors forming an unloaded voltage divider, you can say the voltage at the op-amp's ⊕ input is half that at the junction of the three resistors. And you know that the voltage at the op-amp's ⊕ input must equal Vo, so you can immediately mark the voltage at the junction of the three resistors as equal to ... ?

This simplifies the network you need to analyze. Remember, amplifiers are not all designed to give a gain > 1.0
 

Related to Find the gain in a non-ideal op amp - check my logic

1. What is an op amp and how does it work?

An op amp, short for operational amplifier, is an electronic circuit component commonly used in signal processing and amplification. It typically has two inputs, an inverting and non-inverting, and one output. The output is a function of the difference between the two inputs, with a very high gain or amplification factor.

2. Why is an op amp considered non-ideal?

An op amp is considered non-ideal because it has certain limitations and imperfections that affect its performance. These include input and output impedance, bandwidth, and offset voltage, among others. These imperfections can impact the accuracy and stability of the op amp's output.

3. How do you calculate the gain of a non-ideal op amp?

The gain of a non-ideal op amp can be calculated using the formula A = (Vout/Vin), where A is the gain, Vout is the output voltage, and Vin is the input voltage. However, in non-ideal op amps, the gain may vary depending on the frequency, input impedance, and other factors.

4. What factors affect the gain of a non-ideal op amp?

The gain of a non-ideal op amp can be affected by several factors, including the input and output impedance, open-loop gain, bandwidth, and feedback resistance. These factors can introduce errors and cause the op amp's gain to deviate from its ideal value.

5. How can I check my logic when finding the gain of a non-ideal op amp?

To check your logic when finding the gain of a non-ideal op amp, you can use a circuit simulator or perform calculations using the op amp's transfer function. It is also important to consider the limitations and imperfections of the op amp and how they may affect the gain. Additionally, double-checking your calculations and considering the practical aspects of the circuit can help ensure accurate results.

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