Ideal Opamp Analysis: Determine Vout for Vin = 2V & Vs = 15V

In summary, the ideal op-amp has infinite gain, but in order to amplify input voltages, it requires dc feedback to establish a suitable operating point. This feedback is determined by the feedback resistors only. Without this feedback, small input voltages cannot be amplified due to the always present offset voltage, which would drive the output into saturation. Depending on the opamp, the output can either be pinned at the negative supply rail or saturate close to it. In this specific case, the output would be pinned at -15V since the inverting input is higher than the non-inverting input. This information is relevant for understanding the behavior of op-amps in linear amplifier applications.
  • #1
joel amos
104
0
Given this ideal op-amp, determine Vout if Vin = 2V and Vs = 15 V.
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So I know that the ideal op-amp has infinite gain. As the difference between V+ and V- is substantial (2V), wouldn't that mean the output would theoretically by infinitely large but instead max out at Vs? Am I understanding correctly, or am I way off?
 
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  • #2
Yes - you are right. Each application of opamps as linear amplifiers requires dc feedback. This allows a suitable dc operating point in the mddle region between the supply voltages as well as an amplification factor that is determined by the feedback resistors only.
Otherwise, even very small input voltages cannot be amplified (in case of REAL opamps) because an always present offset voltage will drive the output into saturation.
 
  • #3
Okay, that makes sense. Thanks for the response!
 
  • #4
The way it is drawn the output would be pinned at the -Vs rail, not Vs. The inverting input (-) is higher than the non-inverting input (+) here.
 
  • #5
So it should be -15 V?
 
  • #6
joel amos said:
So it should be -15 V?

Depends on the opamp. Some can drive their output to the rails, and others saturate somewhere close to the rails (within a volt or two). If you are to assume it is an ideal opamp, what do you think?

BTW, is this from a schoolwork assignment? What is the context of the question?
 

1. What is an ideal opamp?

An ideal opamp is an electronic circuit component that has infinite input impedance, infinite open-loop gain, zero output impedance, and infinite bandwidth. It is used in various electronic circuits for amplification, filtering, and other signal processing operations.

2. How is Vout calculated for an ideal opamp?

Vout can be calculated using the formula Vout = A*(Vp-Vn), where A is the open-loop gain of the opamp and Vp and Vn are the voltages at the non-inverting and inverting inputs, respectively. In the case of an ideal opamp, A is assumed to be infinite, so Vout will be equal to the difference between Vp and Vn.

3. What is the significance of Vs in ideal opamp analysis?

Vs represents the supply voltage of the opamp and is an important parameter in determining the output voltage. It limits the maximum and minimum values of Vout based on the power supply voltage and the opamp's saturation levels. In some cases, the supply voltage may also affect the opamp's performance and stability.

4. How is Vin related to Vout in ideal opamp analysis?

Vin is the input voltage applied to the opamp, and it determines the output voltage, Vout. For an ideal opamp, the output voltage will be equal to the input voltage multiplied by the open-loop gain (Vout = A*Vin). This relationship is the basis for many opamp circuits, such as amplifiers and filters.

5. What is the difference between ideal and non-ideal opamp analysis?

In ideal opamp analysis, we assume that the opamp has perfect characteristics, such as infinite gain and bandwidth. However, in reality, opamps have some limitations, such as finite gain, bandwidth, and non-zero output impedance. Non-ideal opamp analysis takes these limitations into account, and the calculations may be more complex compared to ideal opamp analysis.

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