Hi,
According to R5, R6 & R7, U2A output is high & U3b output low only when the voltage at U2a pin 4, V_in satisfies 6v < V_in < 7v (both bounding voltages w.r.t. ground).
Component failure, usually means (for a resistor or capacitor or diode) an open circuit. But what about for a transistor or op-amp or comparator ? If R1, R2,...R7 are each in turn replaced by a 20 Meg resistor (ie open circuit for all practical purposes), U2A out = U3b out for each case. So the failed component cannot be a resistor.
So we're back to my earlier question. In searching for answers I found this:
https://books.google.co.uk/books?id...EwAA#v=onepage&q=op amp failure modes&f=false
For a failed U1a the possible output voltages are: +15v, 0v & -15v w.r.t. ground. Since none of these satisfies the above boundary conditions for V_in, U1a isn't the failed component, and V_in must be 0v.
If U2a has failed then U3b hasn't. The possible output voltages for a failed U2a would be +15V, 0v, -15v w.r.t. ground and the output voltage for U3b will be +15v w.r.t. ground. Since none of these possibilities would cause U4 to turn on, a failed U2a is ruled out.
If U3b has failed then its possible outputs would be +15v, 0v, -15v w.r.t. ground, whereas U2a output would be +15v w.r.t. ground. Two of the possible outputs for a failed U3b, namely 0v & -15v would cause U4 to turn on & light CR1.
Finally, in U4 we have two possible failure modes. First, the LED is busted. In this case the phototransistor will never turn on, which rules out the LED bust. Second, the phototransistor is busted. This leads to two possible outcomes (a) a permanent open circuit between the collector & emitter, and (b) a permanent short circuit between the same. Outcome (b) will result in CR1 lighting.
Hence, the two possible component failures which result in CR1 getting lit with TP1 at 0v w.r.t. ground, are U3b and the phototransistor in optocoupler U4. Whew (sadly there is no emoticon for tired / brain fatigue! I would've loved to put one in here
) !