Timeline for Use of superposition principle for the inverting amplifier
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2018 at 9:35 | history | edited | LvW | CC BY-SA 4.0 |
added 210 characters in body
|
| May 29, 2018 at 9:31 | comment | added | LvW | No - from the beginning I do not know anything about the inverting properties of the circuit. i assume Vi as positiv (against ground) and the same applies to Vout (vs. ground). Both voltages create Vn1 resp. Vn2 at the inverting terminal successively (super position principle). For starting at Aol finite, we set Vn=-Vout/Aol (see EDIT of my answer). | |
| May 29, 2018 at 8:21 | comment | added | Mussé Redi | Shouldn't \$V_{n1}\$ and \$V_{n2}\$ have opposite sign because of the orientation of the current? Or is this opposing sign absorbed into \$V_{out}\$? | |
| May 29, 2018 at 8:18 | comment | added | Mussé Redi | Yes, it is clear to me that \$V_n\$ becomes very small for a very large gain. I may have misstated my initial intention of wanting to derive the second golden rule since it apparently always boils down to the forementioned argument. My main aim for this post is actually to derive the transfer function of an ideal inverted amplifier, starting from the finite gain of an OpAmp and then cranking the gain up to obtain the TF of the ideal OpAmp. | |
| May 29, 2018 at 8:10 | comment | added | LvW | Why Vn=0 ? Answer: As long as the whole circuit operates in its linear amplification region (not saturated) the input voltage at the inverting terminal is always Vn=-(Vout/Aol). Since the open-loop gain of the opamp is very large (1E4...1E5), the voltage Vn is very small (usually in the microvolt range). This small voltage can be neglected in most cases aginst the input voltage Vi and the ouput voltage Vout (usually in the upper millivolt or in the volt range). Hence, for calculation we are allowed to set Vn=0 - and the resulting error is very small. Otherwise, we set Vn=-(Vout/Aol). | |
| May 29, 2018 at 8:02 | comment | added | Mussé Redi | Okay, so now my question is: what is the n-terminal? | |
| May 29, 2018 at 8:01 | comment | added | LvW | For superposition of two voltage sources (here we have two soureces: Vi and Vout) we calculate the circuit parameters in two separate steps: Vn1 is the voltage at the inverting n-terminal as caused by Vi only and Vn2 is the voltage at this input terminal caused by Vout only. Then we add both parts to get the resulting voltage vn. For ideal opams, we may set Vn=0 and for real opams we set Vn=-(Vout/Aol) with Aol=open-loop gain. | |
| May 29, 2018 at 7:54 | comment | added | Mussé Redi | Which voltages refer to \$V_{n1}\$ and \$V_{n2}\$ respectively? | |
| May 29, 2018 at 7:53 | comment | added | Mussé Redi | Why can we set the voltage \$V_n\$ between ground and the inverting terminal to 0? | |
| May 29, 2018 at 6:10 | history | edited | LvW | CC BY-SA 4.0 |
added 1 character in body
|
| May 29, 2018 at 5:47 | history | answered | LvW | CC BY-SA 4.0 |