Find coefficients of op-amp summing circuit

Circuit: V1 (through R1) and V3 (through R3) to the inverting input, R4 from output to inverting input, V2 through R2 to the non-inverting input with R5 to ground. Ideal op-amp assumed.

              R4
      +-----/\/\/-----+
 V1 --R1--+            |
 V3 --R3--+---(-)      |
                 >-----+---- V0
 V2 --R2--+---(+)
          |
          R5
          |
         GND

(i) Coefficients a1, a2, a3

Non-inverting input voltage (voltage divider, no input current):

V+ = V2·R5/(R2 + R5)

KCL at the inverting input, with V− = V+ = Vp:

(V1 - Vp)/R1 + (V3 - Vp)/R3 + (V0 - Vp)/R4 = 0
V0 = -(R4/R1)V1 - (R4/R3)V3 + Vp(1 + R4/R1 + R4/R3)

Substituting Vp:

a1 = -R4/R1
a3 = -R4/R3
a2 = (1 + R4/R1 + R4/R3) · R5/(R2 + R5)
   = [1 + R4/(R1 ‖ R3)] · R5/(R2 + R5)

So V0 = −(R4/R1)V1 + [1 + R4/(R1‖R3)]·[R5/(R2 + R5)]·V2 − (R4/R3)V3.

(ii) R4 short-circuited (R4 = 0)

Then a1 = a3 = 0 and a2 = R5/(R2 + R5). The output is tied directly to the inverting input and the circuit is a voltage follower of V+:

V0 = V2·R5/(R2 + R5)

V1 and V3 have no effect (the op-amp output supplies their currents).

(iii) R4 removed (R4 = ∞)

There is no feedback, so the op-amp works open-loop as a comparator with gain A (very large):

V- = (V1·R3 + V3·R1)/(R1 + R3)
V+ = V2·R5/(R2 + R5)
V0 = +Vsat  if V+ > V-
V0 = -Vsat  if V+ < V-

The output saturates near the supply rails (about ±13 V for ±15 V supply); the linear relation V0 = a1v1 + a2v2 + a3v3 no longer holds.