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Sunday, March 3, 2019

A student designs an electronic sensor that is to be used to switch on a lamp when the light intensity is low. Part of the circuit is shown in Fig. 10.1.


Question 6
A student designs an electronic sensor that is to be used to switch on a lamp when the light intensity is low. Part of the circuit is shown in Fig. 10.1.


Fig. 10.1

(a) State the name of the component labelled X on Fig. 10.1. [1]


(b) On Fig. 10.1, draw the symbols for
(i) two resistors to complete the circuit for the sensing device, [2]

(ii) a relay to complete the circuit for the processing unit. [2]


(c) (i) State the purpose of the relay. [1]

(ii) Suggest why the diode is connected to the output of the operational amplifier
(op-amp) in the direction shown. [2]





Reference: Past Exam Paper – June 2012 Paper 41 & 43 Q10





Solution:
(a) light-dependent resistor


(b)
(i)
two resistors in series between +5 V line and earth
midpoint connected to inverting input of op-amp

(ii)
relay coil between diode and earth
switch between lamp and earth         



(c)
(i)
The relay allows the mains supply to be switched on and off using a low voltage.

(ii)
So that the relay will switch on for one polarity of the output voltage – which is when the output voltage is negative.

2 comments:

  1. Does a protective diode need to be connected across the circuit? Is there back e.m.f when V out is negative? Thanks

    ReplyDelete
    Replies
    1. it could be connected.

      back emf arises when the switch of the relay is closed (due to electromagnetic effects). this occurs whether the output is positive or negative.

      it does not depend on the output itself, BUT rather on the when switch of the relay is closed

      Delete

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