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)

A battery of e.m.f. 6.0 V and negligible internal resistance is connected to three resistors, each of resistance 2.0 kΩ, and a thermistor, as shown in Fig. 9.1.

Question 13

A battery of e.m.f. 6.0 V and negligible internal resistance is connected to three resistors, each of resistance 2.0 kΩ, and a thermistor, as shown in Fig. 9.1.

Fig. 9.1

The thermistor has resistance 2.8 kΩ at 10 °C and resistance 1.8 kΩ at 20 °C.

(a) Calculate the potential

(i) at point A, [1]

(ii) at point B for the thermistor at 10 °C, [2]

(iii) at point B for the thermistor at 20 °C. [1]

(b) The points A and B in Fig. 9.1 are connected to the inputs of an ideal operational amplifier (op-amp), as shown in Fig. 9.2.

Fig. 9.2

The thermistor is warmed from 10 °C to 20 °C.

State and explain the change in the output potential VOUT of the op-amp as the thermistor is warmed. [4]

Reference: Past Exam Paper – November 2015 Paper 41 & 42 Q9

Solution:

(a)

(i) (+) 3.0 V

{Since the resistors have the same resistance, the e.m.f. is equally divided.}

(ii)

{Potential divider equation: V1 = e.m.f. × (R1 / (R1+R2))

The lower resistors are connected to the negative terminal (0 V) of the battery. To find the potential at B, we need to consider the lower resistor.}

potential = 6.0 × {2.0 / (2.0 + 2.8)}

potential = 2.5 V                                 

(iii)

potential = 6.0 × {2.0 / (2.0 + 1.8)}

potential = 3.2 V

(b)

{Recall:

The potential at A (V_A) is fixed at 3 V (as calculated in (a)(i)). It is connected to V^-.

The potential at B (V_B) depends on the temperature of the thermistor. It is connected to V⁺.}

At 10 °C, VA > VB, (since V_B = 2.5 V as calculate in (a)(ii)).

The output voltage V_OUT is -9.0 V (since V^- > V⁺).

At 20 °C, the output voltage VOUT is +9.0 V (as V⁺ is now greater than V^-).

There is a sudden switch (from –9 V to +9 V) when VA = VB.

{As the temperature is increased, V_B would become equal to V_A at some point. It is at this point that the polarity changes.}