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Monday, September 7, 2020

An operational amplifier (op-amp) may be used as part of the processing unit in an electronic sensor.


Question 14
An operational amplifier (op-amp) may be used as part of the processing unit in an electronic sensor.

(a) State three properties of an ideal op-amp. [3]


(b) A comparator circuit incorporating an ideal op-amp is shown in Fig. 9.1.


Fig. 9.1

(i) In one application of the comparator, V2 is kept constant at +1.5 V.
The variation with time t of the potential V1 is shown in Fig. 9.2. The potential V2 is also shown.


Fig. 9.2

On Fig. 9.2, show the variation with time t of the output potential VOUT . [4]


(ii) Two light-emitting diodes (LEDs) R and G are connected to the output of the op-amp in Fig. 9.1 such that R emits light for a longer time than G.

On Fig. 9.1, draw the symbols for the two diodes connected to the output of the op-amp and label the diodes R and G. [3]





Reference: Past Exam Paper – November 2012 Paper 41 & 42 Q9





Solution:
(a) Choose any 3:
Zero output impedance / resistance
Infinite input impedance / resistance
Infinite (open loop) gain
Infinite bandwidth
Infinite slew rate


(b)
(i)
{The output voltage of the comparator can be obtained by
Vout = Ao × (V2 – V1)              where Ao about 105

As the gain A0 is very large, the output of the op-amp will most of the time be saturated, that is, the voltage is either be + 5V or – 5V unless the difference between the two input voltages is very small.

V2 is kept at + 1.5 V.

When the input voltage V2 is greater than V1, the output voltage is positive and equal to + 5 V (as the op-amp is saturated).

Vout = +5V (+ve supply line)


At time t = 0, V2 is greater than V1. So, the output voltage is initially positive.

When the input voltage V2 is less than V1, the output voltage is negative and equal to -5 V (as the op-amp is saturated).

Vout = -5V (-ve supply line)

Thus, the polarity of the output depends on which of the two input voltages is larger.

The output voltage is zero whenever V1 = V2 (from Vout = Ao × (V2 – V1), Vout = 0 V). This is when the polarity of the output changes.}



Graph: square wave                           
Correct cross-over points where V2 = V1
Amplitude 5V                                    
Correct polarity (positive at t=0)       


(ii)
{LED R emits light for a longer time than LED G – from the graph, this corresponds to positive as the positive section is longer. 

Thus, LED R connects when the output is positive. Current flows from positive to negative. So, LED R should point downwards (from the positive output, towards earth).

Also, LED G would point upwards.}



Correct symbol for LED                                            
Diodes connected correctly between Vout and earth
Correct polarity consistent with graph (i)

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