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Op amp circuit simulation problem [Copy link]

Is there something wrong with the voltage and current in this simulation?

20180113_012049.png (75.97 KB, downloads: 0)

20180113_012049.png
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Understand the operation. The 20V at the virtual short same-direction input end is divided by two resistors to get 11.05V. The virtual short reverse input is also 11.05V. The voltage difference between the two ends of the reverse input end is 1.05V. The resistor is 100 ohms and the current at the reverse input end is 10.05ma. There is no problem.  Details Published on 2018-1-15 18:57

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The resistor in the box has a voltage of 1V across its two ends, a resistance of 100Ω, but a current of 10.5mA. According to Ohm's law, do you think there is any problem?
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If you know why the op amp output is 26.8, you will know why the other one is 10.5.
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I don't know what simulation software this is, but it's obviously an incorrect result
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Substituting into the formula, the output voltage is correct
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Using the circuit formula directly, we can indeed get an output voltage of 16.3V, but at the same time, we can also get a voltage of 11.05V at the two input terminals of the op amp. It is this tiny 50mV error that makes the simulation result appear to be inconsistent with Ohm's law.  Details Published on 2018-1-15 08:31
 
 
 
 

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hepen2017 Published on 2018-1-14 23:45 Substituting into the formula, the output voltage is correct
Directly using the circuit formula to calculate, it can indeed get the output voltage is 16.3V, but at the same time, it can also get the voltage of the two input terminals of the op amp is 11.05V. It is this tiny 50mV error that makes the simulation result appear to be inconsistent with Ohm's law
This post is from Analog electronics
 
 
 
 

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Understand the operation. The 20V at the virtual short same-direction input end is divided by two resistors to get 11.05V. The virtual short reverse input is also 11.05V. The voltage difference between the two ends of the reverse input end is 1.05V. The resistor is 100 ohms and the current at the reverse input end is 10.05ma. There is no problem.
This post is from Analog electronics
 
 
 
 

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