__Question 1__**(a)**An ideal operational amplifier (op-amp) has infinite open-loop gain and infinite input resistance (impedance).

State three further properties of an ideal op-amp.
[3]

**(b)**The circuit of Fig. 10.1 is used to detect changes in temperature.

**Fig. 10.1**

The voltmeter has infinite resistance.

The variation with temperature θ of the resistance

*R*of the thermistor is shown in Fig. 10.2.**Fig. 10.2**

**(i)**When the thermistor is at a temperature of 1.0 °C, the voltmeter reads +1.0 V.

Show that, for the thermistor at 1.0 °C, the
potential at A is −0.20 V. [4]

**(ii)**The potential at A remains at −0.20 V.

Determine the voltmeter reading for a thermistor
temperature of 15 °C. [2]

**(c)**The voltmeter reading for a thermistor temperature of 29 °C is 0.35 V.

**(i)**Assuming a linear change of voltmeter reading with change of temperature over the

**(ii)**Suggest why your answers in

**(b)(ii)**and

**(c)(i)**are not the same. [1]

**Reference:**

*Past Exam Paper – November 2014 Paper 41 & 42 Q10*

__Solution 1:__**(a)**

zero output resistance / impedance

infinite bandwidth

infinite slew rate

**(b)**

**(i)**

{from graph:} at 1.0 °C, thermistor resistance is 3.7 kΩ

{for inverting amplifier: voltage gain = V

_{out}/ V_{in}= - R_{f}/ R_{in}}
amplifier gain = –R / 740 = –3700 / 740 (negative
sign essential)

gain = –5.0 C1

{potential at A = V

_{in}= V_{out}/ gain = = 1.0 / –5.0}
potential at A = 1.0 / –5.0 = –0.20 V

**(ii)**

{from graph:} at 15 °C, R = 2.15 kΩ (allow ±0.05 kΩ)

{V

_{out}= - (R_{f}/ R_{in}) V_{in}}
reading = (2150 / 740) × 0.2

reading = 0.58 V (0.59 V → 0.57 V)

**(c)**

**(i)**

{Since they are assumed to vary linearly,
the voltmeter reading is inversely proportional to the temperature (the graph
is a straight line with negative gradient).

29 °C corresponds to 0.35 V

1 °C corresponds to 0.35 × 29

15 °C corresponds to 0.35 × 29 / 15 = 0.68 V}

0.68 V

**(ii)**The resistance (of the thermistor) does not change linearly with temperature