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

V_out = A_o × (V₂ – V₁)              where A_o 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).

V_out = +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).

V_out = -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 V_out = A_o × (V₂ – V₁), V_out = 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 V_out and earth

Correct polarity consistent with graph (i)