A comparator circuit is designed to switch on a mains lamp when the ambient light level reaches a set value.

Question 10

A comparator circuit is designed to switch on a mains lamp when the ambient light level reaches a set value.

An incomplete diagram of the circuit is shown in Fig. 6.1.

Fig. 6.1

(a) (i) A relay is required as part of the output device. This is not shown in Fig. 6.1. Explain why a relay is required. [2]

(ii) On Fig. 6.1, draw the symbol for a relay connected in the circuit as part of the output

device. [2]

(b) Describe the function of

(i) the variable resistor RV, [1]

(ii) the diode D. [1]

(c) State whether the lamp will switch on as the light level increases or as it decreases. Explain your answer. [3]

 [Total: 9]

Reference: Past Exam Paper – June 2017 Paper 41 & 43 Q6

Solution:

(a)

(i)  The op-amp can only deliver a small current but the lamps requires a large current to operate properly.

(ii)

(b)

(i) The variable resistor Rv alters the light intensity at which the lamp would be switched on.

{The LDR and the variable resistor forms a potential divider circuit. That is, some of the 6V of the supply goes to the LDR and some to the variable resistor, depending on the resistance of each.

For example, the resistance of the LDR increases when the light intensity decreases. This results in a p.d. across the LDR. For example, let’s say it is 4V.

If the resistance of the variable resistor is now increased, more voltage would appear across it and so, the p.d. across the LDR decreases.

So, to have a p.d. of 4V again across the LDR, the light intensity on the LDR should further decrease, so that its resistance (and thus, the p.d. across it) increases.}

(ii) The diode is used so that the relay operates for only for one direction of current. This would allow the lamp to operate either for dark conditions or light conditions.

(c)

The resistance of an LDR increases when the light intensity decreases.

When the resistance of the LDR is low (high light level), the p.d. across it is small and so, the potential at V- is big. V- is greater than V+ and so, the output voltage from the op-amp is negative. (Current would flow from earth towards the op-amp, BUT the diode prevents this from happening.)

So, the lamp only comes on when the light level decreases.