You should find Vo/Vs as you say, you don't "find" Vs.Cocoleia said:
berkeman said:You should find Vo/Vs as you say, you don't "find" Vs.
Write the KCL equations for the nodes, and use the properties of an ideal opamp to help simplify the problem. Please show us the KCL equations, and your work toward solving them.
Ok, I will try to simplify that first. But do I have the right idea, would I use the same methods once I simplify?CWatters said:
Let's see your reduced circuit first. It may not be worth writing equations for itCocoleia said:
gneill said:Let's see your reduced circuit first. It may not be worth writing equations for it
gneill said:
gneill said:
I'm not sure what the problem is.gneill said:
Ask yourself if there is any effective feedback that will limit the gain of the circuit. Can any amount of current fed back from the output through the 100 kΩ change the voltages that the op-amp sees at its inputs?Cocoleia said:
It was a bit off track since you were solving for Vs, which should be an independent variable and not dependent on the circuit operation. It's an input. However, you did recognize that there was some kind of problem since it appeared that infinities were springing up.
Isn't the current entering the op amp supposed to be 0A? for both inputsgneill said:
Yes, but feedback is supposed to cause changes in potential differences on components leading to those inputs. Can any amount of current through the "feedback" path make any difference in the potentials at the op-amp inputs in this circuit?Cocoleia said:
If it is too big? I'm not sure I understand, my professor never mentioned feedback in his notes.gneill said:
Feedback is what sets the gain of an op-amp circuit. The so-called open-loop gain of an (ideal) op-amp is infinite. Feedback "closes the loop" and controls the overall gain of the circuit, limiting it to some finite value.Cocoleia said:
How can I explain that is going to be infinity in my answer ? How do I prove this, are there calculations etc ?gneill said:
It's a property of the ideal op-amp. All you have to say is that the potential difference between the inputs is driven directly by Vs. Then refer to the relevant property of the ideal op-amp.Cocoleia said:
Ok, thank you very much !gneill said:
Gain is a measure of how much an opamp amplifies a signal. It is typically expressed in decibels (dB) and is calculated by dividing the output voltage by the input voltage. Opamps are electronic devices that can amplify a signal by a certain amount, known as the gain.
The gain of an opamp circuit can be calculated using the formula: G = -(Rf/Rin), where Rf is the feedback resistor and Rin is the input resistor. This formula assumes an inverting opamp configuration. For non-inverting configurations, the formula is G = 1 + (Rf/Rin).
In an inverting opamp configuration, the input signal is connected to the inverting input terminal, while the non-inverting input terminal is grounded. This results in a negative output voltage. In a non-inverting configuration, the input signal is connected to the non-inverting input terminal, and the output voltage is the same polarity as the input signal.
When selecting an opamp for a specific circuit, factors such as input and output voltage range, gain, bandwidth, and slew rate should be considered. It is also important to ensure that the opamp can handle the desired frequency and power requirements.
Yes, there are limitations to the gain that can be achieved with opamps. These limitations include the power supply voltage, input and output voltage range, and the opamp's internal circuitry. Additionally, high gain can lead to stability issues and increase the chances of noise and distortion in the output signal.