Question

Now, consider the following circuit, but do not construct it yet: 100kΩ 100kΩ BULB 10pF PHOTO CELL R T PUSH BUTTON \ į RELAY The overall purpose of this circuit is to turn on the bulb when the Photocell is placed in the dark and to leave it on until the circuit is reset by depressing the push button. When the output of the op-amp on the right is a positive voltage, the relay closes and the bulb is powered. When is the output of this rightmost op-amp a positive voltage? What must the voltage be on its input? To place this voltage on the input of the rightmost op-amp, what must the voltage on the output of the comparator to the left be?

Answer 1

Answer:

The rightmost op-amp is a voltage-follower. In other words the voltage at its output is same as the voltage at its non-inverting input. This op-amp works like a buffer between comparator stage and the relay. So when the input to rightmost op-amp is positive, its output will also be positive. This voltage must be same as the operating voltage of the relay. For example if the relay coil needs 12V, then input to rightmost op-amp should also be 12V.

To place this voltage on the input of rightmost op-amp, the output voltage of leftmost op-amp should be 0.7v higher. This is so as 0.7V gets dropped across the diode.

Hope this helps.