3 ms·

The way I keep it in my pocket is: The op-amp subtracts the voltage of both inputs, multiplies the result by infinite, and presents it as the output voltage (li

by sasaf5 5y ago

The way I keep it in my pocket is: The op-amp subtracts the voltage of both inputs, multiplies the result by infinite, and presents it as the output voltage (limited by the power supply). When you calculate this with the feedbacks, it makes sense.

dragontamer 5y ago
I never liked this explanation alone. The full explanation is that the + and - terminals are a virtual short. Negative feedback will try to set the + and - terminals to the same voltage but no current flows between the two locations. The amount of current (and the limitations of voltages) are set by the quality of the opamp. Better opamps have closer voltages at faster speeds at better currents. Cheaper opamps will have a bit of error and diverge from the model. The delay can be largely modeled by capacitance over the + and - terminals (among other models of opamp errors) But even a cheap opamp will keep the + and - sides within .05 Volts or so under typical circumstances.
- sasaf5 5y ago
  All true, but I never liked this explanation. How do you use it to analyze a given circuit? With the infinite gain concept you can use a limit to calculate the behavior of any circuit.
  - dragontamer 5y ago
    With virtual short method you set + and - terminals to the same voltage but 0 current between them. This is only true for negative feedback cases but that's the most common case. It is easier in most cases than infinite gain model IMO. Once you have voltage and current of that location figured out, everything else in the circuit is solvable by freshman level analysis.
- exmadscientist 5y ago
  One small trap: current never (in normal cases; yes, I know about the antiparallel diodes and how to deal with/avoid them...) actually flows between the input terminals, it flows to the rails. So you will not, except by sheer luck, get 10nA into the (+) terminal and 10nA out of the (-) terminal. You will get 10nA into the (+) terminal and 4nA into the (-) terminal, both returned to VEE.

by sasaf5 5y ago

dragontamer 5y ago

I never liked this explanation alone. The full explanation is that the + and - terminals are a virtual short. Negative feedback will try to set the + and - terminals to the same voltage but no current flows between the two locations. The amount of current (and the limitations of voltages) are set by the quality of the opamp. Better opamps have closer voltages at faster speeds at better currents. Cheaper opamps will have a bit of error and diverge from the model. The delay can be largely modeled by capacitance over the + and - terminals (among other models of opamp errors) But even a cheap opamp will keep the + and - sides within .05 Volts or so under typical circumstances.

sasaf5 5y ago
All true, but I never liked this explanation. How do you use it to analyze a given circuit? With the infinite gain concept you can use a limit to calculate the behavior of any circuit.
- dragontamer 5y ago
  With virtual short method you set + and - terminals to the same voltage but 0 current between them. This is only true for negative feedback cases but that's the most common case. It is easier in most cases than infinite gain model IMO. Once you have voltage and current of that location figured out, everything else in the circuit is solvable by freshman level analysis.
exmadscientist 5y ago
One small trap: current never (in normal cases; yes, I know about the antiparallel diodes and how to deal with/avoid them...) actually flows between the input terminals, it flows to the rails. So you will not, except by sheer luck, get 10nA into the (+) terminal and 10nA out of the (-) terminal. You will get 10nA into the (+) terminal and 4nA into the (-) terminal, both returned to VEE.