LDR lamp-switching circuit

06/11/2019 Week 6, 6.3, Multisim Session 2, pages 8-11, Activity 1.3 (inc MLQ 1.3, all parts). Note when LDR1 < R2, the lamp does not light. When LDR1 > R2, the lamp lights. This is because of the voltage between them being different in each case. The voltage between them is compared [hence the name 'comparator' applied to this circuit on page 8?] with the voltage between R3 and R4. If the voltage arriving at the positive (+) input of the op-amp is greater when compared with the voltage arriving at the negative (-) input of the op-amp, then the output voltage from the op-amp will be positive. The output voltage will have the same polarity as the diode, and a current will pass through the diode and then through the lamp. As it passes through the lamp, it will illuminate it. NOTE, HOWEVER, that contrary to what previous pages have lead us to expect, when LDR1 is given the value 1400 ohms (not shown here), and the lamp therefore lights up, the output voltage of the op-amp is considerably less than the 12 V supplying the power of the op-amp (assumed supplied through the 'hidden' two terminals of the 3-terminal op-amp). PAGE 29 tells us that: '... The reason for this is that the lamp is drawing 462mA, whereas the op-amp is not intended to supply significant current [so 462 mA is significant current] and therefore power. This has changed the behaviour of the op-amp from the expected amplification characteristics. THIS WILL BE INVESTIGATED FURTHER IN [WEEK 2] SESSION 2... NOTE that although Multism Live continues running, A REAL OP-AMP COULD BE DAMAGED WHEN USED OUTSIDE ITS SPECIFICATION.' [So look out for this in real life!]