Let's figure out the current.
- First off, no current into the + input, therefore no voltage drop across R84.
- The voltage of the + input is 0.
- The voltage of the - input must also be 0.
- No current into - input so the current going through the keyboard resistor string is equal to the current going through R83 and R179 by Kirchoff's Current Law.
There's a trimpot so I will solve for current at both extremes of the trimpot position. First when the pot wiper is at +12V, I = 12/14.3k-12/301k = 0.799 mA. At -12V I = 12/14.3k+12/301k = 0.879 mA. Between those positions, the trimpot can be adjusted to our magic 0.833 mA to give 1/12 V across each 100 ohm resistor.
A question I have looking at this: Why is the high note keyboard voltage 2.66V? We have 32 keys, but 31 resistors between them. V = 31*1/12 = 2.58 V. Someone made a math error and figured 32*1/12 = 2.66 V.
Another question I have looking at this: What is R84 for? Why not connect the + input directly to ground? In our ideal op amp world it works but notice the op amp is a 4558. In the not-so-ideal world this op amp uses bipolar transitors which have a bias current coming out of the + input. So the + input is a little bit above ground in our example. From the datasheet the input bias current is 50 nA. The + input voltage is 50 nA*18k = 0.9 mV. Also from the datasheet the input offset voltage is 1 mV. That means the - input is held very close to ground (roughly 0.9-1 = 0.1 mV) because of R84.
R83 and R179 are 1% to ensure the proper current range. C22 is presumably there for stability while still allowing the output voltage to change quickly enough when new notes are played.
There are plenty more where that came from. Check back for the Rogue, Minimoog, and more.
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