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 = Vout
/ Vin = - Rf / Rin}
amplifier gain = –R / 740 = –3700 / 740 (negative
sign essential)
gain = –5.0 C1
{potential at A = Vin = Vout
/ 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)
{Vout = - (Rf / Rin)
Vin}
Vin}
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
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