Timeline for Virtual ground in a non-inverting ideal op-amp
Current License: CC BY-SA 4.0
7 events
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| Dec 22, 2022 at 7:36 | history | edited | Rohat Kılıç | CC BY-SA 4.0 |
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| Dec 22, 2022 at 7:34 | comment | added | Rohat Kılıç |
@codingliketheresnoyesterday So if there isn't feedback, the voltage at the inverting terminal can be assumed to be 0 or very close to 0 ... Theoretically 0, as it's terminated with a 1k resistor. Practically, since almost all of the op amps have their input bias current flow from pin to outside (i.e. not into the pin) that bias current can generate microvolts, nanovolts, or even millivolts across that 1k resistor (close to 0). ... when a voltage is applied to the non-inverting terminal? regardless of the non-inv terminal voltage, we should say. Applying supply voltage only is enough.
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| Dec 22, 2022 at 7:25 | comment | added | Designalog | @codingliketheresnoyesterday it is close to 0 for the very reason that you connected it to ground via s resistor. | |
| Dec 22, 2022 at 7:25 | comment | added | Glenn Willen | The opamp does not directly control the voltage at the input terminals. The only thing it controls is the output terminal. So if there is no feedback, the input terminals will have exactly the voltages they look like they have. The op-amp without feedback has as much impact on the input terminals as a brick, or a ham sandwich. (I.e. none whatsoever.) | |
| Dec 22, 2022 at 7:17 | vote | accept | codingliketheresnoyesterday | ||
| Dec 22, 2022 at 7:17 | comment | added | codingliketheresnoyesterday | I have no idea how I have never realized this until now. So if there isn't feedback, the voltage at the inverting terminal can be assumed to be 0 or very close to 0 when a voltage is applied to the non-inverting terminal? | |
| Dec 22, 2022 at 7:08 | history | answered | Rohat Kılıç | CC BY-SA 4.0 |