Timeline for Solving an Op-amp circuit
Current License: CC BY-SA 3.0
25 events
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| Jun 11, 2020 at 15:10 | history | edited | CommunityBot |
Commonmark migration
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| Feb 1, 2014 at 19:40 | vote | accept | user29568 | ||
| Feb 1, 2014 at 15:10 | comment | added | Alfred Centauri | @user29568, see the update to my answer. | |
| Feb 1, 2014 at 14:23 | history | edited | user29568 | CC BY-SA 3.0 |
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| Feb 1, 2014 at 13:17 | comment | added | user29568 | @AlfredCentauri I still don't see the bottom loop, do you mean the loop \$v_+\$ connected to \$V_B\$ then connected to the voltage source and then the resistor and finally \$V_A\$. Is that considered a loop even with the op-amp? And when I do I still don't get your equation. | |
| Feb 1, 2014 at 12:30 | comment | added | Alfred Centauri | @user29568, I simply wrote a KVL equation 'round the bottom-most loop and that equation is correct. Obviously, if the resistor were replaced with a wire, we would have \$V_B = V_A + 2V\$ so your equation is incorrect. | |
| Feb 1, 2014 at 7:40 | comment | added | user29568 | @AlfredCentauri You must be using node equations, because the only "loop" I can see is the two \$V_A\$ connected by a virtual short circuit--except if I am looking at this in the wrong way. And If it's node equation(which use KVL) then isn't it \$V_B=V_A-2-i_11\text{k}\Omega\$ | |
| Jan 31, 2014 at 21:21 | comment | added | Alfred Centauri | @user29568, \$V_B\$ does not equal 2V because of the voltage source. Denote the current from left to right through the 1k resistor connected to node A as \$i_1\$. Then, by KVL, we have: \$V_B = V_A - i_1 1k\Omega + 2V \$ | |
| Jan 31, 2014 at 20:47 | comment | added | user29568 | @AlfredCentauri As node A and the negative terminal ofthe op-Amp are connected directly they have the same voltage so \$V_A=v_{-}=v_+\$ and \$V_B=2\$ because of the voltage source. Clearly, I have asked to many comments and taken to much of your time; 50 min past like a second. So, I don't want to take anymore of your time thanks :). | |
| Jan 31, 2014 at 20:44 | history | edited | user29568 | CC BY-SA 3.0 |
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| Jan 31, 2014 at 20:40 | comment | added | user29568 | @AlfredCentauri I am going to add a new schematic with new labels. Do you mind me asking you a few questions? | |
| Jan 31, 2014 at 20:36 | comment | added | Alfred Centauri | @user29568, according my calculations, \$V_O = -1.5V\$ and \$v_+ = v_- = -2V\$ | |
| Jan 31, 2014 at 20:35 | comment | added | user29568 | Also, how did you get, \$v_- = V_O + i_x \cdot 1k\Omega\$ by KVL. I am clearly too lost :((( | |
| Jan 31, 2014 at 20:25 | comment | added | user29568 | \$V_O\$ is what I meant. How can \$V_B=V_O\$ if there is a resistor between them? | |
| Jan 31, 2014 at 20:23 | comment | added | Alfred Centauri | @user29568, you'll have to answer that. I don't see a node voltage labelled \$v_0\$ on either diagram. By the way, as you've labelled things in the 2nd diagram, \$V_B = V_O\$. | |
| Jan 31, 2014 at 20:22 | comment | added | user29568 | I see...then what is \$v_0\$? IF it is meant to be the voltage across point B and the ground then the voltage is 2V. | |
| Jan 31, 2014 at 20:20 | comment | added | Alfred Centauri | @user29568, according to the first diagram, \$V_O\$ is not the voltage across the resistor connected to ground; the voltage across that resistor is \$v_+\$. The voltage of the output terminal (reference to ground of course) is \$V_O\$. | |
| Jan 31, 2014 at 20:16 | comment | added | user29568 | Is \$v_0\$ representing the voltage across the resistor connected to the ground? If yes, then the node I have labeled as \$V_0\$ should have voltage \$V_o\$ no? | |
| Jan 31, 2014 at 20:13 | comment | added | Alfred Centauri | @user29568, no, why do you think that? | |
| Jan 31, 2014 at 19:58 | comment | added | user29568 | @AlfredCentauri Isn't \$v_+=v_o=v_-\$ | |
| Jan 31, 2014 at 19:55 | comment | added | Alfred Centauri | Regarding your 2nd schematic: \$v_-\$ is not equal to \$V_O\$. By KVL, \$v_- = V_O + i_x \cdot 1k\Omega\$ | |
| Jan 31, 2014 at 19:44 | history | edited | user29568 | CC BY-SA 3.0 |
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| Jan 31, 2014 at 14:38 | answer | added | Alfred Centauri | timeline score: 3 | |
| Jan 31, 2014 at 13:57 | comment | added | Pyxzure | Nodal analysis is still useful. 'Ground' is just referencing a point to be 0V. Instead of calling it 'ground', you could call it V1 V2 and it is still the same. You can start by making an equation for each node, then continue from there. | |
| Jan 31, 2014 at 13:17 | history | asked | user29568 | CC BY-SA 3.0 |