The output potential VOUT from an operational amplifier is to be monitored using an output device.

Question 7

The output potential VOUT from an operational amplifier is to be monitored using an output device.

The output VOUT can be either +5 V or −5 V.

(a) On Fig. 10.1, draw a circuit for the output device that consists of two light-emitting diodes B and G.

Diode B alone is to emit light when VOUT is +5 V. Diode G alone is to emit light when

VOUT is −5 V.

Fig. 10.1

[3]

(b) On Fig. 10.2, draw a circuit of the output device that consists of a relay and a diode such that a high-power lamp is switched on only when VOUT is −5 V.

Fig. 10.2

[4]

Reference: Past Exam Paper – November 2015 Paper 43 Q10

Solution:

(a)

correct LED symbol                           

separately connected between VOUT and earth with opposite polarities

diode B ‘pointing’ from VOUT to earth  

(ignore protective resistors)

 

{Current flows from higher potential to lower potential.

Diode B is to be used when V_out = + 5 V. Current would flow from V_out to earth (downward) since V_out is at a higher potential. So, diode B should point downwards.

When V_out = -5 V, current flows from earth to V_out – that is, upwards. So, diode G should point upwards.}

(b)

diode in VOUT line

diode ‘pointing’ towards VOUT from earth

relay coil connected between VOUT and earth

switch connected across lamp

{Since V_out = - 5 V, current would flow from earth to V_out. So, the diode should point towards the output to allow the current to flow.

Another diode may be added in parallel to the relay coil (to protect the op-amp) as when a current is switched off in the electromagnet (relay coil), a back e.m.f. is induced in the coil that could damage the op-amp. To prevent this current from flowing to the op-amp, the diode should be downwards.}